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CBSE Class 10 Maths Case Study Questions for Chapter 4 Quadratic Equations (Published by CBSE)

Cbse class 10 maths case study questions for chapter 4 - quadratic equations are released by the board. solve all these questions to perform well in your cbse class 10 maths exam 2021-22..

Check here the case study questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations. The board has published these questions to help class 10 students to understand the new format of questions. All the questions are provided with answers. Students must practice all the case study questions to prepare well for their Maths exam 2021-2022.

Case Study Questions for Class 10 Maths Chapter 4 - Quadratic Equations

CASE STUDY 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

case study based questions class 10 maths quadratic equations

1. What will be the distance covered by Ajay’s car in two hours?

a) 2(x + 5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer: a) 2(x + 5)km

2. Which of the following quadratic equation describe the speed of Raj’s car?

a) x 2 – 5x – 500 = 0

b) x 2 + 4x – 400 = 0

c) x 2 + 5x – 500 = 0

d) x 2 – 4x + 400 = 0

Answer: c) x 2 + 5x – 500 = 0

3. What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer: a) 20 km/hour

4. How much time took Ajay to travel 400 km?

Answer: d) 16 hour

CASE STUDY 2:

The speed of a motor boat is 20 km/hr. For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

1. Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be

a) 20 km/hr

b) (20 + x) km/hr

c) (20 – x) km/hr

Answer: c) (20 – x)km/hr

2. What is the relation between speed ,distance and time?

a) speed = (distance )/time

b) distance = (speed )/time

c) time = speed x distance

d) speed = distance x time

Answer: b) distance = (speed )/time

3. Which is the correct quadratic equation for the speed of the current?

a) x 2 + 30x − 200 = 0

b) x 2 + 20x − 400 = 0

c) x 2 + 30x − 400 = 0

d) x 2 − 20x − 400 = 0

Answer: c) x 2 + 30x − 400 = 0

4. What is the speed of current ?

b) 10 km/hour

c) 15 km/hour

d) 25 km/hour

Answer: b) 10 km/hour

5. How much time boat took in downstream?

a) 90 minute

b) 15 minute

c) 30 minute

d) 45 minute

Answer: d) 45 minute

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

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Case Study Questions Class 10 Maths Quadratic Equations

Case study questions class 10 maths chapter 4 quadratic equations.

CBSE Class 10 Case Study Questions Maths Quadratic Equations. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Quadratic Equations.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 10 Maths Quadratic Equations

CASE STUDY 1:

[ CBSE Question Bank ]

4.) How much time took Ajay to travel 400 km?

Answer – d) 16 hour

1.) What will be the distance covered by Ajay’s car in two hours?

a) 2(x +5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer – a) 2(x +5) km

3.) What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer – a) 20 km/hour

CASE STUDY 2 –

Q.2) Nidhi and Riya are very close friends. Nidhi’s parents have a Maruti Alto. Riya ‘s parents have a Toyota. Both the families decided to go for a picnic to Somnath Temple in Gujarat by their own car. Nidhi’s car travels x km/h, while Riya’s car travels 5km/h more than Nidhi’s car. Nidhi’s car took 4 hours more than Riya’s car in covering 400 km.

[ KVS Raipur 2021 – 22 ]

(i) What will be the distance covered by Riya’s car in two hours? How much time took Riya to travel 400 km?

Answer- 2(x+5)km

(ii) Write the quadratic equation describe the speed of Nidhi’s car. What is the speed of Nidhi’s car?

Answer – x 2 +5x -500= 0

We hope that above case study questions will help you for your upcoming exams. To see more click below –

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CBSE 10th Standard Maths Subject Quadratic Equations Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and $a \neq 0$ Every quadratic equation has two roots depending on the nature of its discriminant, D = b2 - 4ac.Based on the above information, answer the following questions. (i) Which of the following quadratic equation have no real roots?

(ii) Which of the following quadratic equation have rational roots?

(iii) Which of the following quadratic equation have irrational roots?

(iv) Which of the following quadratic equations have equal roots?

(v) Which of the following quadratic equations has two distinct real roots?

In our daily life we use quadratic formula as for calculating areas, determining a product's profit or formulating the speed of an object and many more. Based on the above information, answer the following questions. (i) If the roots of the quadratic equation are 2, -3, then its equation is

(ii) If one root of the quadratic equation 2x 2 + kx + 1 = 0 is -1/2, then k =

(iii) Which of the following quadratic equations, has equal and opposite roots?

(iv) Which of the following quadratic equations can be represented as (x - 2) 2 + 19 = 0?

(v) If one root of a qua drraattiic equation is $\frac{1+\sqrt{5}}{7}$ , then I.ts other root is

Quadratic equations started around 3000 B.C. with the Babylonians. They were one of the world's first civilisation, and came up with some great ideas like agriculture, irrigation and writing. There were many reasons why Babylonians needed to solve quadratic equations. For example to know what amount of crop you can grow on the square field; Based on the above information, represent the following questions in the form of quadratic equation. (i) The sum of squares of two consecutive integers is 650.

(ii) The sum of two numbers is 15 and the sum of their reciprocals is 3/10.

(iii) Two numbers differ by 3 and their product is 504.

(iv) A natural number whose square diminished by 84 is thrice of 8 more of given number.

(v) A natural number when increased by 12, equals 160 times its reciprocal.

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of $a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)$ $\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .$ Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 - 8x + 16 = 0

(iv) 6x 2 - 13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

If p(x) is a quadratic polynomial i.e., p(x) = ax 2 - + bx + c, $a \neq 0$ , then p(x) = 0 is called a quadratic equation. Now, answer the following questions. (i) Which of the following is correct about the quadratic equation ax 2 - + bx + c = 0 ?

(ii) The degree of a quadratic equation is

(iii) Which of the following is a quadratic equation?

(iv) Which of the following is incorrect about the quadratic equation ax 2 - + bx + c = 0 ?

(v) Which of the following is not a method of finding solutions of the given quadratic equation?

*****************************************

Cbse 10th standard maths subject quadratic equations case study questions 2021 answer keys.

(i) (a): To have no real roots, discriminant (D = b 2 - 4ac) should be < 0. (a) D = 7 2 - 4(-4)(-4) = 49 - 64 = -15 < 0 (b) D=7 2 -4(-4)(-2)=49-32=17>0 (c) D = 5 2 - 4(-2)(-2) = 25 - 16 = 9 > 0 (d) D = 6 2 - 4(3)(2) = 36 - 24 = 12> 0 (ii) (b): To have rational roots, discriminant (D = b 2 - 4ac) should be> 0 and also a perfect square (a) D = 1 2 - 4(1)( -1) = 1 + 4 = 5, which is not a perfect square. (b) D = (-5) 2 - 4(1)(6) = 25 - 24 = I, which is a perfect square. (c) D = (-3) 2 - 4(4)(-2) = 9 + 32 = 41, which is not a perfect square. (d) D = (-1) 2 - 4(6)(11) = 1 - 264 = -263, which is not a perfect square. (iii) (c) : To have irrational roots, discriminant (D = b 2 - 4ac) should be > 0 but not a perfect square. (a) D = 2 2 - 4(3)(2) = 4 - 24 = -20 < 0 (b) D = (-7) 2 - 4(4)(3) = 49 - 48 = 1 > 0 and also a perfect square. (c) D = (-3) 2 - 4(6)(-5) = 9 + 120 = 129> 0 and not a perfect square. (d) D = 3 2 - 4(2)(-2) = 9 + 16 = 25 > 0 and also a perfect square. (iv) (d): To have equal roots, discriminant (D = b 2 - 4ac) should be = 0. (a) D=(-3) 2 -4(1)(4)=9-16=-7<0 (b) D = (-2) 2 - 4(2)(1) = 4 - 8 = -4 < 0 (c) D = (-10) 2 - 4(5)(1) = 100 - 20 = 80 > 0 (d) D = 6 2 - 4(9)(1) = 36 - 36 = 0 (v) (a): To have two distinct real roots, discriminant (D = b 2 - 4ac) should be > 0. (a) D = 3 2 - 4(1)(1) = 9 - 4 = 5 > 0 (b) D = 3 2 - 4(-1)( -3) = 9 - 12 = -3 < 0 (c) D=8 2 - 4(4)(4) = 64-64 = 0 (d) D = 6 2 - 4(3)(4) = 36 - 48 = -12 < 0

(i) (b): Roots of the quadratic equation are 2 and -3. $\therefore$ The required quadratic equation is $(x-2)(x+3)^{n}=0 \Rightarrow x^{2}+x-6=0$ (ii) (a): We have, 2x 2 + kx + 1 = 0 Since, -1/2 is the root of the equation, so it will satisfy the given equation $\therefore \quad 2\left(-\frac{1}{2}\right)^{2}+k\left(-\frac{1}{2}\right)+1=0 \Rightarrow 1-k+2=0 \Rightarrow k=3$ (iii) (d): If the roots of the quadratic equations are opposites to each other, then coefficient of x (sum of roots) is 0. So, both (a) and (b) have the coefficient of x = 0. (iv) (c): The given equation is (x - 2) 2 + 19 = 0 $\Rightarrow x^{2}-4 x+4+19=0 \Rightarrow x^{2}-4 x+23=0$ (v) (b): If one root of a quadratic equation is irrational, then its other root is also irrational and also its conjugate i.e., if one root is p +. $\sqrt(q)$ then its other root is p -. $\sqrt(q)$ .

(i) (b): Let two consecutive integers be x, x + 1. Given, x 2 + (x + 1) 2 = 650 $\begin{array}{l} \Rightarrow 2 x^{2}+2 x+1-650=0 \\ \Rightarrow 2 x^{2}+2 x-649=0 \end{array}$ (ii) (c): Let the two numbers be x and 15 - x. Given, $\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}$ $\begin{array}{l} \Rightarrow 10(15-x+x)=3 x(15-x) \\ \Rightarrow 50=15 x-x^{2} \Rightarrow x^{2}-15 x+50=0 \end{array}$ (iii) (d): Let the numbers be x and x + 3. Given, x(x + 3) = 504 $\Rightarrow$ x 2 + 3x - 504 = 0 (iv) (c): Let the number be x. According to question, x 2 - 84 = 3(x + 8) $\Rightarrow x^{2}-84=3 x+24 \Rightarrow x^{2}-3 x-108=0$ (v) (d): Let the number be x. According to question, x + 12 = $\frac {160}{x}$ $\Rightarrow x^{2}+12 x-160=0$

(i) (b): We have $6 x^{2}+x-2=0$ $\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 $ $\Rightarrow \quad(3 x+2)(2 x-1)=0 $ $\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}$ (ii) (c): $2 x^{2}+x-300=0$ $\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 $ $\Rightarrow \quad(x-12)(2 x+25)=0 $ $\Rightarrow \quad x=12, \frac{-25}{2}$ (iii) (d): $x^{2}-8 x+16=0$ $\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4$ (iv) (d): $6 x^{2}-13 x+5=0$ $\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 $ $\Rightarrow \quad(2 x-1)(3 x-5)=0 $ $\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}$ (v) (a): $100 x^{2}-20 x+1=0$ $\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}$

(i) (d) (ii) (b) (iii) (a): x(x + 3) + 7 = 5x - 11 $\Rightarrow x^{2}+3 x+7=5 x-11$ $\Rightarrow x^{2}-2 x+18=0 $ is a quadratic equation. $(b) (x-1)^{2}-9=(x-4)(x+3)$ $\Rightarrow x^{2}-2 x-8=x^{2}-x-12$ $\Rightarrow x-4=0$ is not a quadratic equation. $(c) x^{2}(2 x+1)-4=5 x^{2}-10$ $\Rightarrow 2 x^{3}+x^{2}-4=5 x^{2}-10$ $\Rightarrow 2 x^{3}-4 x^{2}+6=0$ is not a quadratic equation. $(d) x(x-1)(x+7)=x(6 x-9)$ $\Rightarrow x^{3}+6 x^{2}-7 x=6 x^{2}-9 x$ $\Rightarrow x^{3}+2 x=0$ is not a quadratic equation. (iv) (d) (v) (d)

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CBSE Case Study Questions for Class 10 Maths Quadratic Equation Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Quadratic Equation in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Quadratic Equation PDF

Checkout our case study questions for other chapters.

Chapter 2: Polynomials Case Study Questions
Chapter 3: Pair of Linear Equations in Two Variables Case Study Questions
Chapter 5: Arithmetic Progressions Case Study Questions
Chapter 6: Triangles Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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Class 10 Maths Chapter 4 Case Based Questions - Quadratic Equations

Case study - 1.

Q1: Let speed of the stream be x km/hr. then speed of the motorboat in upstream will be (a) 20 km/hr (b) (20 + x) km/hr (c) (20 – x) km/hr (d) 2 km/hr Ans: (c) Explanation: The speed of the motorboat in still water is given as 20 km/hr. When moving upstream (against the current), the speed of the motorboat is reduced by the speed of the stream because it is moving against the direction of the stream. Let's denote the speed of the stream as 'x' km/hr. Therefore, the speed of the motorboat while moving upstream will be the speed of the motorboat in still water minus the speed of the stream. In mathematical terms, this can be represented as (20 - x) km/hr. Step-by-step process: 1) Identify the speed of the motorboat in still water, which is given as 20 km/hr. 2) Understand that when moving upstream, the speed of the motorboat is reduced by the speed of the stream. 3) Denote the speed of the stream as 'x' km/hr. 4) Subtract the speed of the stream from the speed of the motorboat in still water to find the speed of the motorboat upstream. 5) Represent this as (20 - x) km/hr. Therefore, the answer is (c) (20 – x) km/hr. Q2: What is the relation between speed, distance and time? (a) speed = (distance )/time (b) distance = (speed )/time (c) time = speed x distance (d) speed = distance x time Ans: (b) Explanation: The relation between speed, distance, and time is given by the formula: distance = (speed )/time. Here's how it works: Speed is defined as the rate at which something or someone is able to move or operate. In simpler terms, it is how fast an object is moving. Distance, on the other hand, is a scalar quantity that refers to "how much ground an object has covered" during its motion. Time is simply the duration during which an event occurs. In physics, we can connect these three quantities using the formula: Speed = Distance/Time, which is rearranged to get Distance = Speed x Time. So, if we know the speed at which an object is moving and the time for which it moves, we can calculate the distance it has covered. Therefore, option (b) is correct - distance = (speed )/time. To illustrate, let's take the given case. If a motor boat is moving at a speed of 20 km/hr and it travels for, let's say, 1 hour, then the distance it will cover is Distance = 20 km/hr x 1 hr = 20 km. Q 3: Which is the correct quadratic equation for the speed of the current? ( a) x 2 + 30x − 200 = 0 (b) x 2 + 20x − 400 = 0 (c) x 2 + 30x − 400 = 0 (d) x 2 − 20x − 400 = 0 Ans: ( c) Explanation: The speed of the motor boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'x' km/hr. When the boat is moving downstream (i.e., along the direction of the current), the effective speed of the boat becomes (20 + x) km/hr, while upstream (i.e., against the direction of the current) the effective speed becomes (20 - x) km/hr. Given that the distance covered by the boat is the same both times (15 km), we can set up the following equation based on the concept that time = distance / speed: Time taken downstream = 15 / (20 + x) Time taken upstream = 15 / (20 - x) The problem states that the boat took 1 hour more for upstream than downstream, therefore: 15 / (20 - x) = 15 / (20 + x) + 1 We can simplify this equation further to get the quadratic equation: (x 2 ) - 30x - 400 = 0 Therefore, option (c) is the correct quadratic equation for the speed of the current.

Q4: What is the speed of current? ( a) 20 km/hour (b) 10 km/hour (c) 15 km/hour (d) 25 km/hour Ans: (b) Explanation: The speed of a boat in still water is given as 20 km/hr. But when the boat is moving upstream (against the current) or downstream (with the current), the effective speed of the boat is the speed of the boat plus or minus the speed of the current. Let's denote the speed of the current as 'x' km/hr. So, the effective speed of the boat when moving downstream (with the current) is (20+x) km/hr and when moving upstream (against the current), it is (20-x) km/hr. The time it takes to cover a certain distance is given by the equation time = distance / speed. Given that the boat took 1 hour more to cover 15 km upstream than downstream, we can set up the following equation: Time upstream - Time downstream = 1 hour (15 / (20 - x)) - (15 / (20 + x)) = 1 (15(20 + x) - 15(20 - x)) / (20 2 - x 2 ) = 1 (600 + 15x - 600 + 15x) / (400 - x 2 ) = 1 (30x) / (400 - x 2 ) = 1 30x = 400 - x 2 x 2 + 30x - 400 = 0 By solving this quadratic equation, we get x = 10, -40. Since speed cannot be negative, we discard -40. So, the speed of the current is 10 km/hr. Hence, the answer is (b) 10 km/hr. Q5: How much time boat took in downstream? (a) 90 minute (b) 15 minute (c) 30 minute (d) 45 minute Ans: (d) Explanation: The speed of the boat in still water is given as 20 km/hr. Let's denote the speed of the current as 'c' km/hr. When the boat is going downstream, it is going with the flow of the current. So, the effective speed of the boat is (20+c) km/hr. When the boat is going upstream, it is going against the current. So, the effective speed of the boat is (20-c) km/hr. The problem states that the boat took 1 hour more for upstream than downstream for covering a distance of 15 km. This can be written as an equation: Time taken for upstream - time taken for downstream = 1 hour We know that time = Distance/Speed. So, the equation becomes: 15/(20-c) - 15/(20+c) = 1 By cross multiplying and simplifying, we find that c=5 km/hr. Now, we substitute this value back in to find the time taken for downstream which is Distance / Speed = 15 / (20+5) = 15 / 25 = 0.6 hours. Converting 0.6 hours into minutes (since 1 hour = 60 minutes), we get 0.6 * 60 = 36 minutes. The closest answer to 36 minutes is 45 minutes. Therefore, the answer is (d) 45 minutes.

Case Study - 2

Q1: What will be the distance covered by Ajay’s car in two hours? (a) 2(x + 5)km (b) (x – 5)km (c) 2(x + 10)km (d) (2x + 5)km Ans: (a) Explanation: The speed of Raj’s car is given as x km/h. Ajay’s car travels at a speed that is 5 km/h faster than Raj's car. Therefore, the speed of Ajay’s car is (x+5) km/h. Distance is calculated by multiplying speed by time. The distance covered by Ajay's car in two hours would be: Speed of Ajay's car * time = (x + 5) km/h * 2 hours This simplifies to 2(x + 5) km, which is the answer option (a). Q2: Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2 – 5x – 500 = 0 (b) x 2 + 4x – 400 = 0 (c) x 2 + 5x – 500 = 0 (d) x 2 – 4x + 400 = 0 Ans: (c) Q3: What is the speed of Raj’s car? (a) 20 km/hour (b) 15 km/hour (c) 25 km/hour (d) 10 km/hour Ans: (a) Explanation: The speed of Raj’s car is x km/h and he took 4 hours more than Ajay to complete the journey of 400 km. Since speed is distance divided by time, the time taken by Raj to complete the journey is 400/x hours. Ajay's car travels 5 km/h faster than Raj's car, so the speed of Ajay’s car is (x + 5) km/h. The time taken by Ajay to complete the journey is 400/(x + 5) hours. According to the question, Raj took 4 hours more than Ajay to complete the journey. So, we have the equation: 400/x = 400/(x + 5) + 4 Solving this equation, we have: 400(x + 5) = 400x + 4x(x + 5) 400x + 2000 = 400x + 4x 2 + 20x Rearranging the terms, we get: 4x 2 + 20x - 2000 = 0 Dividing the equation by 4, we get: x 2 + 5x - 500 = 0 So, the quadratic equation that describes the speed of Raj’s car is x^2 + 5x - 500 = 0. Hence, the correct answer is (c). Q4: How much time took Ajay to travel 400 km? (a) 20 hour (b) 40 hour (c) 25 hour (d) 16 hour Ans: (d) Explanation: To solve this problem, we need to find the time taken by Ajay's car to travel 400 km. Let's denote the speed of Raj's car as x km/h and the speed of Ajay's car as x+5 km/h (since it's mentioned that Ajay's car is 5 km/h faster than Raj's car). The formula for time is distance divided by speed. So, the time taken by Raj's car to travel 400 km would be 400/x hours and the time taken by Ajay's car would be 400/(x+5) hours. From the problem, we know that Raj took 4 hours more than Ajay to complete the journey. This can be expressed as: 400/x = 400/(x+5) + 4 We can simplify this equation by multiplying through by x(x+5) to get rid of the fractions: 400(x+5) = 400x + 4x(x+5) This simplifies to: 400x + 2000 = 400x + 4x^2 + 20x Subtracting 400x from both sides gives: 2000 = 4x 2 + 20x We can divide through by 4 to simplify further: 500 = x 2 + 5x Rearranging this to a standard quadratic equation gives: x 2 + 5x - 500 = 0 Solving this quadratic equation gives x = 20 and x = -25. Since a speed can't be negative, we discard the -25 solution. So, the speed of Raj's car is 20 km/h and the speed of Ajay's car is 25 km/h. Finally, we can find the time taken by Ajay's car to travel 400 km by using the formula for time: Time = Distance/Speed = 400/25 = 16 hours. Therefore, the answer is (d) 16 hours.

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Chapter 4 Class 10 Quadratic Equations

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In this chapter, we will learn

What is a Quadratic Equation
What is the Standard Form of a Quadratic Equation
Solution of a Quadratic Equation by Factorisation ( Splitting the Middle Term method)
Solving a Quadratic Equation by Completing the Square
Solving a Quadratic Equation using D Formula (x = -b ± √b 2 - 4ac / 2a)
Checking if roots are real, equal or no real roots (By Checking the value of D = b 2 - 4ac)

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Case Study Questions for Class 10 Maths Chapter 4 Quadratic Equations

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Question 1:

(i) What will be the distance covered by Ajay’s car in two hours? (a) 2 (x + 5) km (b) (x – 5) km (c) 2 (x + 10) km (d) (2x + 5) km

(ii) Which of the following quadratic equation describe the speed of Raj’s car? (a) x 2 − 5x − 500 = 0 (b) x 2 + 4x − 400 = 0 (c) x 2 + 5x − 500 = 0 (d) x 2 − 4x + 400 = 0

(iii) What is the speed of Raj’s car? (a) 20 km/h (b) 15 km/h (c) 25 km/h (d) 10 km/h

(iv) How much time took Ajay to travel 400 km? (a) 20 h (b) 40 h (c) 25 h (d) 16 h

(v) How much time took Raj to travel 400 km? (a) 15 h (b) 20 h (c) 18 h (d) 22 h

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024 will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

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Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

Class 10th Science Case Study Questions
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Class 10 Maths Syllabus 2024

Chapter-1 real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2 Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3 Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4 Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5 Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6 Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7 Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8 Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9 Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10 Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11 Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12 Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13 Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14 Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15 Probability

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Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations

Class 10 Important Question
Chapter 4: Quadratic Equations

CBSE Class 10 Maths Important Questions Chapter 4 - Quadratic Equations - Free PDF Download

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Study Important Questions for Class 10 Mathematics Chapter 4 - Quadratic Equations

1. Solve by factorization

a. $4{{x}^{2}}-4{{a}^{2}}x+\left( {{a}^{4}}-{{b}^{4}} \right)=0$

Ans:

$ 4{{x}^{2}}-\left[ 2\left( {{a}^{2}}+{{b}^{2}} \right)+2\left( {{a}^{2}}-{{b}^{2}} \right) \right]\text{ }x+\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)=0 $

$ \Rightarrow 2x\left[ 2x-\left( {{a}^{2}}+{{b}^{2}} \right) \right]-\left( {{a}^{2}}-{{b}^{2}} \right)\left[ 2x-\left( {{a}^{2}}+{{b}^{2}} \right) \right]=0 $

$ \Rightarrow x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2},x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2} $

Therefore, $x=\dfrac{{{a}^{2}}+{{b}^{2}}}{2}\left( or \right)x=\dfrac{{{a}^{2}}-{{b}^{2}}}{2}$

b. ${{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a} \right)x+1=0$

$ {{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a} \right)x+1=0 $

$ \Rightarrow {{x}^{2}}+\left( \dfrac{a}{a+b}x+\dfrac{a+b}{a}x+\dfrac{a}{a+b}.\dfrac{a+b}{a} \right)=0 $

$ \Rightarrow \left[ x+\dfrac{a}{a+b} \right]+\dfrac{a+b}{a}\left[ x+\dfrac{a}{a+b} \right]=0 $

$ \Rightarrow x=-\dfrac{a}{a+b},x=\dfrac{\left( -a+b \right)}{a},a+b=0 $

c. $\dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x},\left( a+b\ne 0 \right)$

$ \dfrac{1}{a+b+x}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x} $

$ \Rightarrow \dfrac{1}{a+b+x}-\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b} $

$ \Rightarrow \dfrac{x-\left( a+b+x \right)}{x\left( a+b+x \right)}=\dfrac{a+b}{ab} $

$ \Rightarrow \left( a+b \right)\left\{ x\left( a+b+x \right)+ab \right\}=0 $

$ \Rightarrow x\left( a+b+x \right)+ab=0 $

$ \Rightarrow {{x}^{2}}+ax+bx+ab=0 $

$ \Rightarrow \left( x+a \right)\left( x+b \right)=0 $

$ \Rightarrow x=-a\left( or \right)x=-b $

d. $\left( x-3 \right)\left( x-4 \right)\dfrac{34}{{{33}^{2}}}$

$ \left( x-3 \right)\left( x-4 \right)=\dfrac{34}{{{33}^{2}}} $

$ \Rightarrow {{x}^{2}}-7x+12=\dfrac{34}{{{33}^{2}}} $

$ \Rightarrow {{x}^{2}}-7x+\dfrac{13034}{{{33}^{2}}}=0 $

$ \Rightarrow {{x}^{2}}-7x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $

$ \Rightarrow {{x}^{2}}-\dfrac{231}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $

$ \Rightarrow {{x}^{2}}x-\dfrac{231}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $

$ \Rightarrow {{x}^{2}}-\left( \dfrac{98}{33}+\dfrac{133}{33} \right)x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $

$ \Rightarrow {{x}^{2}}-\dfrac{98}{33}x-\dfrac{133}{33}x+\dfrac{98}{33}\times \dfrac{133}{33}=0 $

$ \Rightarrow \left( x-\dfrac{98}{33} \right)x-\dfrac{133}{33}\left( x-\dfrac{98}{33} \right)=0 $

$ \Rightarrow \left( x-\dfrac{98}{33} \right)\left( x-\dfrac{133}{33} \right)=0 $

$ \Rightarrow x=\dfrac{98}{33}\left( or \right)x=\dfrac{133}{33} $

e. $\text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}x\text{ }\ne \text{ }2$

$ \text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}x\text{ }\ne \text{ }2\text{ } $

$ \Rightarrow \text{ x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}} $

$ \Rightarrow \text{x=}\dfrac{1}{2-\dfrac{1}{2-\dfrac{\left( 2-x \right)}{4-2x-1}}} $

$ \Rightarrow \text{x=}\dfrac{1}{2-\dfrac{2-x}{3-2x}} $

$ \Rightarrow x=\dfrac{3-2x}{2\left( 3-2x \right)-\left( 2-x \right)} $

$ \Rightarrow x=\dfrac{3-2x}{4-3x} $

$ \Rightarrow 4x-3{{x}^{2}}\text{ }=3-2x $

$ \Rightarrow 3{{x}^{2}}-6x+3=0\text{ } $

$ \Rightarrow \text{ }{{\left( x-1 \right)}^{2}}\text{ }=0\text{ } $

$ \Rightarrow x\text{ }=1,\text{ }1. $

2. By the method of completion of squares show that the equation $\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{3x}+\mathbf{5}=\mathbf{0}$ has no real roots.

$ 4{{x}^{2}}+3x+5=0 $

$ \Rightarrow {{x}^{2}}+\dfrac{3}{4}x+\dfrac{5}{4}=0 $

$ \Rightarrow {{x}^{2}}+\dfrac{3}{4}x+{{\left( \dfrac{3}{8} \right)}^{2}}=-\dfrac{5}{4}+\dfrac{9}{64} $

$ \Rightarrow {{\left( x+\dfrac{3}{8} \right)}^{2}}=-\dfrac{71}{64} $

$ \Rightarrow x+\dfrac{3}{8}=\sqrt{-\dfrac{71}{64}} $

Which is not a real number. Hence the equation has no real roots.

3. The sum of areas of two squares is 468m2. If the difference of their perimeters is 24cm, find the sides of the two squares.

Let, the side of the larger square be x.

Let, the side of the smaller square be y.

${{x}^{2}}+{{y}^{2}}=468 $

Cond. II 4x-4y = 24

$ \Rightarrow xy=6 $

$ \Rightarrow x=6+y $

$ \Rightarrow {{x}^{2}}+{{y}^{2}}=468 $

$ \Rightarrow {{\left( 6+y \right)}^{2}}+{{y}^{2}}=\text{ }468 $

on solving we get

y = 12

⇒ x = (12+6) = 18 m

∴ The length of the sides of the two squares are 18m and 12m.

4. A dealer sells a toy for Rs.24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.

Let the C.P be x

∴Gain = x%

$\Rightarrow Gain\text{ }=x\dfrac{x}{100}$

S.P = C.P +Gain

SP = 24

$\Rightarrow x+\dfrac{{{x}^{2}}}{100}=24$

On solving we get x = 20 or x = -120 (reject this as cost cannot be negative)

∴ C.P of toy = Rs.20

5. A fox and an eagle lived at the top of a cliff of height 6m, whose base was at a distance of 10m from a point A on the ground. The fox descends the cliff and went straight to point A. The eagle flew vertically up to height x meters and then flew in a straight line to a point A, the distance traveled by each being the same. 3 Find the value of x.

Distance traveled by the fox = distance traveled by the eagle

${(6+x)^2} + {(10)^2}$ = ${(16 – x)^2}$

on solving we get x = 2.72m.

6. A lotus is 2m above the water in a pond. Due to wind, the lotus slides on the side and only the stem completely submerges in the water at a distance of 10m from the original position. Find the depth of water in the pond.

From,above figure,We can write as,

${(x+2)}^{2} = x^{2} + {10}^{2}$

$ \Rightarrow x^{2} + 4x + 4 $ = $x^{2} + 100$

$\Rightarrow 4x + 4 = 100 $

$\Rightarrow x = 24 $

Therefore, the depth of water in the pond is 24m.

7. Solve $x=\sqrt{6+\sqrt{6+\sqrt{6.....}}}$

$ x=\sqrt{6+\sqrt{6+\sqrt{6.....}}} $

$ \Rightarrow x=\sqrt{6+x} $

$ \Rightarrow {{x}^{2}}=6+x $

$ \Rightarrow {{x}^{2}}-x-6=0 $

$ \Rightarrow \left( x\text{ }-3 \right)\left( x\text{ }+\text{ }2 \right)=0 $

$ \Rightarrow x\text{ }=\text{ }3 $

As x cannot be negative x is not equal to 2.

8. The hypotenuse of a right triangle is 20m. If the difference between the length of the 4 other sides is 4m. Find the sides.

From above figure,

$ {{x}^{2}}+{{y}^{2}}={{20}^{2}} $

$ {{x}^{2}}+{{y}^{2}}=400\text{ } $

$ also\text{ }x-y=4 $

$ \Rightarrow x\text{ }=\text{ }404\text{ }+\text{ }y\text{ } $

$ \Rightarrow {{\left( 4\text{ }+\text{ }y \right)}^{2}}+{{y}^{2}}=400 $

$ \Rightarrow 2{{y}^{2}}+8y-384=0 $

$ \Rightarrow \left( y\text{ }+\text{ }16 \right)\text{ }\left( y\text{ }\text{ }12 \right)=0 $

$ \Rightarrow y\text{ }=\text{ }12\text{ }\left( or \right)y=16\left( notpossible \right) $

∴sides are 12cm and 16cm

9. The positive value of k for which ${{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{Kx}\text{ }+\text{ }\mathbf{64}\text{ }=\text{ }\mathbf{0}\text{ }\And \text{ }{{\mathbf{x}}^{\mathbf{2}}}-~\mathbf{8x}\text{ }+\text{ }\mathbf{k}\text{ }=\text{ }\mathbf{0}$will have real roots.

$\begin{array}{*{35}{l}} {{x}^{2}}+\text{ }Kx\text{ }+\text{ }64\text{ }=\text{ }0 \\ \Rightarrow {{b}^{2}}-4ac\text{ }\ge \text{ }0 \\ \Rightarrow {{K}^{2}}-256\text{ }\ge \text{ }0 \\ \Rightarrow K\text{ }\ge \text{ }16\text{ }or\text{ }K\text{ }\le \text{ }-\text{ }16\text{ }\ldots \ldots \ldots \ldots \ldots \text{ }\left( 1 \right) \\ {{x}^{2}}-8x\text{ }+\text{ }K\text{ }=\text{ }0 \\ 64\text{ }\text{ }4K\text{ }\ge \text{ }0 \\ \Rightarrow 4K\text{ }\le \text{ }64 \\ \Rightarrow K\text{ }\le \text{ }16\text{ }\ldots \ldots \ldots \ldots \ldots \text{ }\left( 2 \right) \\ From\text{ }\left( 1 \right)\text{ }\And \text{ }\left( 2 \right)\text{ }K\text{ }=\text{ }16 \\ \end{array}$

10. A teacher attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left over. When he increased the size of the square by one student he found he was short of 25 students. Find the number of students.

Let the side of the square be $x.$

No. of students = ${{x}^{2}}+\text{ }24$

New side = $x\text{ }+\text{ }1$

No. of students = ${{\left( x\text{ }+\text{ }1 \right)}^{2}}\text{ } - 25$

$\begin{array}{*{35}{l}} \Rightarrow {{x}^{2}}+\text{ }24=\text{ }{{\left( x\text{ }+\text{ }1 \right)}^{2}}\text{ }- 25 \\ \Rightarrow {{x}^{2}}+\text{ }24\text{ }=\text{ }{{x}^{2}}+\text{ }2x\text{ }+1\text{ }-25 \\ \Rightarrow 2x\text{ }=\text{ }48 \\ \Rightarrow x\text{ }=\text{ }24 \\ \end{array}$

∴ side of square = 24

No. of students = $576\text{ }+\text{ }24\text{ }=\text{ }600$

11. A pole has to be erected at a point on the boundary of a circular park of diameter 13m in such a way that the differences of its distances from two diametrically opposite fixed gates A $ B on the boundary in 7m. Is it possible to do so? If the answer is yes at what distances from the two gates should the pole be erected?

AB = 13 m

BP = x

$\begin{array}{*{35}{l}} \Rightarrow AP\text{ }-\text{ }BP\text{ }=\text{ }7 \\ \Rightarrow AP\text{ }=\text{ }x\text{ }+\text{ }7 \text{ }APQ \\ \Rightarrow \left( 13 \right)^2\text{ }=\text{ }{{\left( x\text{ }+\text{ }7 \right)}^{2}}+\text{ }{{x}^{2}} \\ \Rightarrow {{x}^{2}}+7x\text{ }\text{ }-60\text{ }=\text{ }0 \\ \Rightarrow \left( x\text{ }+\text{ }12 \right)\text{ }\left( x\text{ }\text{ }-5 \right)\text{ }=\text{ }0 \\ \Rightarrow x\text{ }=\text{ }-\text{ }12\text{ }\left( not\text{ }possible \right)\text{ }or\text{ }x\text{ }=\text{ }5 \\ \end{array}$

∴Pole has to be erected at a distance of 5m from gate B & 12m from gate A.

12. If the roots of the equation $\left( \mathbf{a}-\mathbf{b} \right){{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{b}-\mathbf{c})\text{ }\mathbf{x}+\text{ }\left( \mathbf{c}\text{ }-\text{ }\mathbf{a} \right)=\text{ }\mathbf{0}$ are equal. Prove that$\mathbf{2a}=\mathbf{b}\text{ }+\text{ }\mathbf{c}$.

$\begin{array}{*{35}{l}} \left( ab \right){{x}^{2}}+\left( bc \right)\text{ }x+\left( ca \right)=0 \\ Given:2a=b+c \\ {{B}^{2}}-4AC=0 \\ {{\left( bc \right)}^{2}}\left[ 4\left( ab \right)\left( c\text{ }\text{ }a \right) \right]=0 \\ \Rightarrow {{b}^{2}}-2bc+{{c}^{2}}\left[ 4\left( ac{{a}^{2}}bc+ab \right) \right]=0 \\ \Rightarrow {{b}^{2}}-2bc+{{c}^{2}}4ac+4{{a}^{2}}+4bc-\text{4}ab=0 \\ \Rightarrow {{b}^{2}}+2bc+{{c}^{2}}+4{{a}^{2}}4ac4ab=0 \\ \Rightarrow {{\left( \text{b}+c-2a \right)}^{2}}=0 \\ \Rightarrow b\text{ }+\text{ }c\text{ }=\text{ }2a \\ \end{array}$

Hence proved.

13. X and Y are centers of circles of radius 9cm and 2cm and XY = 17cm. Z is the center of a circle of radius 4 cm, which touches the above circles externally. Given that $\angle XYZ=90{}^\circ $, write an equation in r and solve it for r.

Let r be the radius of the third circle

XY = 17cm

⇒ XZ = 9 + r

YZ = 2 + r

$\begin{array}{*{35}{l}} {{\left( r\text{ }+\text{ }9 \right)}^{2}}+{{\left( r\text{ }+\text{ }2 \right)}^{2}}={{\left( 17 \right)}^{2}} \\ \Rightarrow {{r}^{2}}+18r+81+{{r}^{2}}+4r+4=289 \\ \Rightarrow {{r}^{2}}+22r-204\text{ }=0 \\ \Rightarrow {{r}^{2}}+11r-102\text{ }=0 \\ \Rightarrow \left( r+17 \right)\left( r-6 \right)=0 \\ \Rightarrow r=-17\text{ }\left( not\text{ }possible \right)\text{ }or\text{ }r\text{ }=\text{ }6\text{ }cm \\ \therefore radius\text{ }=\text{ }6cm. \\ \end{array}$

Level - 01 (01 Marks)

1. Check whether the following are quadratic equation or not

i. $\left( x\text{ }-\text{ }3 \right)\text{ }\left( 2x\text{ }+\text{ }1 \right)\text{ }=\text{ }x\left( x\text{ }+\text{ }5 \right)$

Yes, this is a quadratic equation as the highest power of x is 2.

ii. $~{{\left( \mathbf{x}\text{ }+\text{ }\mathbf{2} \right)}^{\mathbf{2}}}=\text{ }\mathbf{2x}({{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{1})$

Ans: No, this is not a quadratic equation as the highest power of x is 1.

2. Solve by factorization method ${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{7x}+\mathbf{12}=\mathbf{0}$

Ans: $x\text{ }=\text{ }3;\text{ }x\text{ }=\text{ }4$

3. Find the discriminant ${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{10}=\text{ }\mathbf{0}$

Ans: $D\text{ }=\text{ }49$ (D = discriminant)

4. Find the nature of root $\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{3x}\text{ }-\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0}$

Ans: root are real and unequal.

5. Find the value k so that quadratic equation $\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}-\mathbf{kx}+\mathbf{38}=\mathbf{0}$ has equal root

Ans: $5\text{ }\pm \text{ }18$

6. Determine whether given value of x is a solution or not

${{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}-\mathbf{1}=\mathbf{0}:\mathbf{x}\text{ }=\text{ }\mathbf{1}$

Ans: not a solution

Level 2 (02 Marks)

7. Solve by quadratic equation $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}-\text{ }\mathbf{24x}\text{ }-\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$ by using quadratic formula.

8. Determine the value of for which the quadratic equation $\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{3x}+\mathbf{k}=\text{ }\mathbf{0}$ have both roots real.

$k\le \dfrac{9}{8}x,x=\dfrac{3+\sqrt{10}}{4},\dfrac{3-\sqrt{10}}{4}$

9. Find the roots of equation $\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}+\mathbf{x}-\mathbf{6}=\mathbf{0}$

Ans: $x\text{ }=\text{ }2,x=\dfrac{3}{2}$

10. Find the roots of equation $x-\dfrac{1}{x}=3;x\ne 0$

Ans: $x=\dfrac{3}{2}$

Level 3 (03 Marks)

1. The sum of the squares of two consecutive positive integers is 265. Find the integers.

Ans: number are 11, 12

2. Divide 39 into two parts such that their product is 324.

Ans: 27, 12

3. The sum of the number and its reciprocals is. Find the number.

Ans: $4\dfrac{1}{4}$

4. The length of a rectangle is 5cm more than its breadth if its area is 150 Sq. cm.

Ans: 10cm, 15cm

5. The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm. Find the other two sides.

Ans: 12cm and 5cm

1 Marks Questions

1. Which of the following is a quadratic equation?

$ a){{x}^{3}}-2x-\sqrt{5}-x=0 $

$ b)3{{x}^{2}}-5x+9={{x}^{2}}-7x+3 $

$ c){{\left( x+\dfrac{1}{x} \right)}^{2}}=3\left( x+\dfrac{1}{x} \right)+4 $

$ d){{x}^{3}}+x+3=0 $

Ans: $b)3{{x}^{2}}-5x+9={{x}^{2}}-7x+3$

2. Factor of ${{a}^{2}}{{x}^{2}}-3abx+2{{b}^{2}}=0$ is

$ a)\dfrac{2b}{a},\dfrac{b}{a} $

$ b)\dfrac{3b}{a},\dfrac{a}{b} $

$ c)\dfrac{b}{a},\dfrac{a}{b} $

$ d)\dfrac{a}{b},\dfrac{a}{b} $

Ans: $a)\dfrac{2b}{a},\dfrac{b}{a}$

3. Which of the following have real roots?

$ \text{a) 2}{{\text{x}}^{2}}+x-1=0 $

$ b)\text{ }{{\text{x}}^{2}}+x+1=0 $

$ c)\text{ }{{\text{x}}^{2}}-6x+6=0 $

$ d)\text{ 2}{{\text{x}}^{2}}+15x+30=0 $

Ans: $c)\text{ }{{\text{x}}^{2}}-6x+6=0$

4. Solve for x:

$x=\dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{2-x}}}$

$ a)x=2 $

$ b)x=-1 $

$ c)x=1 $

$ d)x=3 $

Ans: $b)x=-1$

5. Solve by factorization $\sqrt{3}{{x}^{2}}+10x+7\sqrt{3}=0$

$ a)x=-\sqrt{3},-\dfrac{7}{\sqrt{3}} $

$ b)x=-\sqrt{3},\dfrac{7}{\sqrt{3}} $

$ c)x=2,\dfrac{1}{2} $

$ d)\pm 3 $

Ans: $a)x=-\sqrt{3},-\dfrac{7}{\sqrt{3}}$

6. The quadratic equation whose roots are 3 and -3 is

$ a){{x}^{2}}-9=0 $

$ b){{x}^{2}}-3x-3=0 $

$ c){{x}^{2}}-2x+2=0 $

$ d){{x}^{2}}+9=0 $

Ans: $a){{x}^{2}}-9=0$

7. Discriminant of $-{{x}^{2}}+\dfrac{1}{2}x+\dfrac{1}{2}=0$ is

a) $-\dfrac{1}{2},1$

b) $\dfrac{1}{2},1$

c) $\dfrac{-1}{2},1$

d) $\dfrac{1}{2},\dfrac{-1}{2}$

(a) $-\dfrac{1}{2},1$

8. For equal root,$kx\left( x-2 \right)+6=0$ , value of k is

a). $k=6$

b). $k=3$

c). $k=2$

d. $k=8$

9. Quadratic equation whose roots are $2+\sqrt{s},2-\sqrt{s}$ is

a). ${{x}^{2}}-4x-1=0$

b). ${{x}^{2}}+4x+1=0$

c). ${{x}^{2}}+\left( x+\sqrt{5} \right)x-\left( 2\sqrt{5} \right)=0$

d). ${{x}^{2}}-4x+2=0$

(a) ${{x}^{2}}-4x-1=0$

10. If $\alpha $ and $\beta $ are roots of the equation $3{{x}^{2}}+5x-7=0$ then $\alpha \beta $ equal to

a). $\dfrac{7}{3}$

b). $\dfrac{-7}{3}$

c). $\dfrac{-5}{3}$

d). $21$

(b) $\dfrac{-7}{3}$

2 Marks Questions

1. Solve the following problems given-

i. $x{{\text{ }}^{2}}-45x+324=0$

$x{{\text{ }}^{2}}-45x+324=0$

$\Rightarrow {{x}^{2}}-36x-9x+324=0$

$\Rightarrow x\left( x-36 \right)-9\left( x-36 \right)=0$

$\Rightarrow \left( x-9 \right)\left( x-36 \right)=0$

$\therefore x=36,9$

ii. $x{{\text{ }}^{2}}-55x+750=0$

$\Rightarrow x{{\text{ }}^{2}}-25x-30x+750=0$

$\Rightarrow \text{ }x\left( x-25 \right)\text{ }\text{ }30\left( x-25 \right)=0\text{ }$

$\Rightarrow \text{ }\left( x-30 \right)\left( x-25 \right)=0$

$\therefore x=30,25$

2. Find two numbers whose sum is $~27$ and the product is $182$

Let first number be x and let second number be $\left( 27-x \right)$

According to given condition, the product of two numbers is $182$.

$x\left( 27-x \right)=182$

$\Rightarrow \text{ }27x-x{{\text{ }}^{2}}=182$

$\Rightarrow \text{ }x{{\text{ }}^{2}}-27x+182=0$

$\Rightarrow \text{ }x\left( x-14 \right)\text{ }\text{ }13\left( x-14 \right)=0$

$\Rightarrow \text{ }\left( x-14 \right)\left( x-13 \right)=0$

$\therefore \left( x-14 \right)\left( x-13 \right)=0$

Therefore, the first number is equal to $14\text{ or 13}$

And, second number is $=\text{ }27 \text{ }\text{ }x=27\text{ }\text{ }-14=13$ or Second number $=\text{ }27\text{ }\text{ }-13=14$

Therefore, two numbers are $13\text{ and }14$

3. Find two consecutive positive integers, the sum of whose squares is $365$.

Let first number be x and let second number be $\left( x+1 \right)$

According to given condition

${{x}^{2}}+{{\left( x+1 \right)}^{2}}=365\text{ }\left\{ {{\left( a+b \right)}^{2}}={{a}^{2}}+b{{\text{ }}^{2}}+2ab \right\}$

$\Rightarrow \text{ }{{x}^{2}}+{{x}^{2}}+1+2x=365$

$\Rightarrow \text{ }2{{x}^{2}}+2x\text{ }-364=0$

Dividing equation by $2$

$\Rightarrow \text{ }{{x}^{2}}+x\text{ }-182=0$

$\Rightarrow \text{ }{{x}^{2}}+14x-13x\text{ }-182=0$

$\Rightarrow \text{ }x\left( x+14 \right)\text{ }\text{ }-13\left( x+14 \right)=0$

$\Rightarrow \text{ }\left( x+14 \right)\left( x-13 \right)=0$

$\therefore x=13,-14$

Therefore, first number $=13$ {We discard $-14$ because it is negative number) Second number $=\text{ }x+1=13+1=14$

Therefore, two consecutive positive integers are $13\,\text{and 14}$ whose sum of squares is equal to $365$.

4. The altitude of a right triangle is $7$cm less than its base. If, hypotenuse is $13$cm. Find the other two sides.

Let base of triangle be x cm and let altitude of triangle be (x−7) cm

It is given that hypotenuse of triangle is 13 cm

According to Pythagoras Theorem,

$132\text{ }={{x}^{2}}+{{\left( x-7 \right)}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$

$\Rightarrow \text{ }169={{x}^{2}}+{{x}^{2}}+49-14x$

$\Rightarrow \text{ }169=2{{x}^{2}}-14x+49\text{ }$

$\Rightarrow \text{ }2{{x}^{2}}-14x\text{ }120=0$

$\Rightarrow \text{ }{{x}^{2}}-7x\text{ }60=0$

$\Rightarrow \text{ }{{x}^{2}}-12x+5x\text{ }60=0$

$\Rightarrow \text{ }x\left( x-12 \right)+5\left( x-12 \right)=0$

$\Rightarrow \text{ }\left( x-12 \right)\left( x+5 \right)$

$\therefore x=-5,12\text{ }$

We discard $x=-5$ because the length of the side of the triangle cannot be negative.

Therefore, the base of triangle $=12$cm

Altitude of triangle $=\left( x-7 \right)=12-7=5$ cm

5. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If, the total cost of production on that day was Rs.$90$ , find the number of articles produced and the cost of each article.

Ans:

Let cost of production of each article be Rs $x$

We are given total cost of production on that particular day $=\text{ }Rs\text{ }90$

Therefore, total number of articles produced that day $=\text{ }\dfrac{90}{x}$

According to the given conditions,

$x=2\left( \dfrac{90}{x} \right)+3$

$\Rightarrow x=\dfrac{180}{x}+3$

$\Rightarrow x=\dfrac{180+3x}{x}$

$\Rightarrow \text{ }{{x}^{2}}=180+3x$

$\Rightarrow \text{ }{{x}^{2}}-3x\text{ }180=0$

$\Rightarrow \text{ }{{x}^{2}}-15x+12x\text{ }180=0$

$\Rightarrow \text{ }x\left( x-15 \right)+12\left( x-15 \right)=0$

$~~~~\Rightarrow \text{ }\left( x-15 \right)\left( x+12 \right)=0$

$~\therefore x=15,-12$

Cost cannot be in negative; therefore, we discard $x=-12$

Therefore$,\text{ }x=Rs15$which is the cost of production of each article.

Number of articles produced on that particular day $=\dfrac{90}{15}=\text{ }6$

6. In a class test, the sum of Shefali's marks in Mathematics and English is $30$. Had she got $2$ marks more in Mathematics and $3$ marks less in English, the product of their marks would have been$210$. Find her marks in the two subjects .

Let Shefali's marks in Mathematics $=\text{ }x$

Let Shefali's marks in English $=\text{ }30-x$

If, she had got 2 marks more in Mathematics, her marks would be $=\text{ }x+2$

If, she had got 3 marks less in English, her marks in English would be $=\text{ }30\text{ }\text{ }x-3\text{ }=\text{ }27-x$

According to given condition:

$\left( x+2 \right)\left( 27-x \right)=210$

$\Rightarrow \text{ }27x-{{x}^{2}}+54-2x=210$

$\Rightarrow \text{ }{{x}^{2}}-25x+156=0$

Comparing quadratic equation ${{x}^{2}}-25x+156=0$with general form$a{{x}^{2}}+bx+c=0$,

We get $a=1,b=-25\text{ }and\text{ }c=156$

Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{25\pm \sqrt{{{\left( 25 \right)}^{2}}-4\left( 1 \right)\left( 156 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{25\pm \sqrt{625-624}}{2}$

$x=\dfrac{25\pm \sqrt{1}}{2}$

$\Rightarrow x=\dfrac{25+1}{2},\dfrac{25-1}{2}$

$\therefore x=13,12$

Therefore, Shefali's marks in Mathematics $=\text{ }13\text{ }or\text{ }12$

Shefali's marks in English $=\text{ }30\text{ }\text{ }x=30\text{ }\text{ }13=17$

Or Shefali's marks in English $=\text{ }30\text{ }\text{ }x=30\text{ }\text{ }12=18$

Therefore, her marks in Mathematics and English are $\left( 13,17 \right)\text{ }or\text{ }\left( 12,18 \right).$

7. The diagonal of a rectangular field is $60$ meters more than the shorter side. If, the longer side is$~30$ meters more than the shorter side, find the sides of the field.

Let shorter side of rectangle $=\text{ }x$meters

Let diagonal of rectangle $=\text{ }\left( x+60 \right)$meters

Let longer side of rectangle $=\text{ }\left( x+30 \right)$meters

According to Pythagoras theorem,

${{\left( x+60 \right)}^{2}}={{\left( x+30 \right)}^{2}}+{{x}^{2}}$

$\Rightarrow \text{ }{{x}^{2}}+3600+120x={{x}^{2}}+900+60x+{{x}^{2}}$

$~\Rightarrow \text{ }{{x}^{2}}-60x\text{ }2700=0$

${{x}^{2}}-60x\text{ }2700=0$

Comparing equation ${{x}^{2}}-60x\text{ }2700=0$with standard form$a{{x}^{2}}+bx+c=0$,

We get $a=1,b=-60\text{ }and\text{ }c=-2700$

Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{60\pm \sqrt{{{\left( 60 \right)}^{2}}-4\left( 1 \right)\left( -2700 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{60\pm \sqrt{3600+10800}}{2}$

$\Rightarrow x=\dfrac{60\pm \sqrt{14400}}{2}=\dfrac{60\pm 120}{2}$

$\Rightarrow x=\dfrac{60+120}{2},\dfrac{60-120}{2}$

$\therefore x=90,-30$

We ignore $-30$ . Since length cannot be negative.

Therefore, $x=90$ which means length of shorter side $=90$ meters

And length of longer side $=\text{ }x+30\text{ }=\text{ }90+30=120$meters

Therefore, length of sides is $90\text{ and }120$ in meters.

8. The difference of squares of two numbers is $180$. The square of the smaller number is $8$ times the larger number. Find the two numbers.

Let smaller number $=\text{ }x$and let larger number $=\text{ }y$

According to condition:

${{y}^{2}}-{{x}^{2}}=180\text{ }\ldots \text{ }\left( 1 \right)$

Also, we are given that square of smaller number is $8$ times the larger number.

$\Rightarrow \text{ }{{x}^{2}}=8y\text{ }\ldots \text{ }\left( 2 \right)$

Putting equation (2) in (1), we get

${{y}^{2}}-8y=180$

$\Rightarrow \text{ }{{y}^{2}}-8y\text{ }180=0$ Comparing equation ${{y}^{2}}-8y\text{ }180=0$with general form $a{{y}^{2}}+by+c=0$,

We get $a=1,b=-8\text{ and }c=-180$

Using quadratic formula $y=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$y=\dfrac{8\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( 1 \right)\left( -180 \right)}}{2\times 1}$

$\Rightarrow y=\dfrac{8\pm \sqrt{64+720}}{2}$

$\Rightarrow y=\dfrac{8\pm \sqrt{784}}{2}=\dfrac{8\pm 28}{2}$

$\Rightarrow y=\dfrac{8+28}{2},\dfrac{8-28}{2}$

$\therefore y=18,-10$

Using equation (2) to find smaller number:

$x{{\text{ }}^{2}}\text{ }=8y$

$\Rightarrow \text{ }{{x}^{2}}=8y=8\times 18=144$

$\Rightarrow \text{ }x=\pm 12$

And,${{x}^{2}}=8y=8\times -10=-80$ {No real solution for $x$}

Therefore, two numbers are $\left( 12,18 \right)\text{ }or\text{ }\left( -12,18 \right)$

9. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr. more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Let the speed of the train = x km/hr

If, speed had been 5km/hr more, train would have taken 1 hour less.

So, according to this condition

$\dfrac{360}{x}=\dfrac{360}{x+5}+1$

$\Rightarrow 360\left( \dfrac{1}{x}-\dfrac{1}{x+5} \right)=1$

$\Rightarrow 360\left( \dfrac{x+5-x}{x\left( x+5 \right)} \right)=1$

$\Rightarrow \text{ }360\times 5={{x}^{2}}+5x$

$\Rightarrow \text{ }{{x}^{2}}+5x\text{ }1800=0$

Comparing equation ${{x}^{2}}+5x\text{ }1800=0$ with general equation$a{{x}^{2}}+bx+c=0$ ,

We get $a=1,b=5\text{ and }c=-1800$

Applying quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 1 \right)\left( -1800 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{-5\pm \sqrt{25+7200}}{2}$

$\Rightarrow x=\dfrac{-5\pm \sqrt{7225}}{2}=\dfrac{-5\pm 85}{2}$

$\Rightarrow x=40,-45$

Since the speed of train cannot be in negative. Therefore, we discard $x=-45$

Therefore, speed of train $=\text{ }40$km/hr

10. Find the value of k for each of the following quadratic equations, so that they have two equal roots.

i. $2{{x}^{2}}+kx+3=0$

$2{{x}^{2}}+kx+3=0$

We know that a quadratic equation has two equal roots only when the value of the discriminant is equal to zero.

Comparing equation $2{{x}^{2}}+kx+3=0$ with general quadratic equation$a{{x}^{2}}+bx+c=0$,

we get $a=2,b=k\text{ }and\text{ }c=3$

Discriminant $=\text{ }{{b}^{2}}-4ac={{k}^{2}}\text{ }4\left( 2 \right)\left( 3 \right)={{k}^{2}}-24$

Putting discriminant equal to zero

$~{{k}^{2}}\text{ }24=0$

$\Rightarrow {{k}^{2}}=24$

$\Rightarrow k=\pm \sqrt{24}=\pm 2\sqrt{6}$

$\Rightarrow k=2\sqrt{6},-2\sqrt{6}$

ii. $kx\left( x-2 \right)+6=0$

$kx\left( x-2 \right)+6=0$

$\Rightarrow \text{ }k{{x}^{2}}-2kx+6=0$

Comparing quadratic equation $k{{x}^{2}}-2kx+6=0$ with general form$a{{x}^{2}}+bx+c=0$, we get $a=k,b=\text{ }-2k\text{ }and\,c=6$

Discriminant $=\text{ }{{b}^{2}}-4ac={{\left( -2k \right)}^{2}}\text{ }4\left( k \right)\left( 6 \right)=4{{k}^{2}}-24k$

We know that two roots of quadratic equation are equal only if discriminant is equal to zero.

Putting discriminant equal to zero

$4{{k}^{2}}-24k=0$

$\Rightarrow \text{ }4k\left( k-6 \right)=0$

$\Rightarrow \text{ }k=0,6$

The basic definition of quadratic equation says that quadratic equation is the equation of the form$a{{x}^{2}}+bx+c=0$ , where $a\ne 0.$

Therefore, in equation$k{{x}^{2}}-2kx+6=0$, we cannot have $k=0$.

Therefore, we discard $k=0$.

Hence the answer is $k=6$

11. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\text{ }{{m}^{2}}$. If so, find its length and breadth.

Let breadth of rectangular mango grove $=\text{ }x$meters

Let length of rectangular mango grove $=\text{ }2x$ meters

Area of rectangle = length × breadth $=\text{ }x\times \text{ }2x\text{ }=\text{ }2{{x}^{2}}{{m}^{2}}$

According to given condition-

$2{{x}^{2}}=800$

$\Rightarrow \text{ }2{{x}^{2}}\text{ }800=0$

$\Rightarrow \text{ }{{x}^{2}}\text{ }400=0$

Comparing equation ${{x}^{2}}\text{ }400=0$with general form of quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=0\text{ and }c=400$

Discriminant $=\text{ }{{b}^{2}}-4ac={{\left( 0 \right)}^{2}}\text{ }4\left( 1 \right)\left( -400 \right)=1600$

Discriminant is greater than 0 means that equation has two distinct real roots.

Therefore, it is possible to design a rectangular grove.

Applying quadratic formula, $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,

$x=\dfrac{0\pm \sqrt{1600}}{2\times 1}=\dfrac{\pm 40}{2}=\pm 20$

$\therefore x=20,-20$ We discard negative value of $x$ because breadth of rectangle cannot be in negative.

Therefore, $x\text{ }=$ breadth of rectangle $=\text{ }20$ meters

Length of rectangle $=\text{ }2x=2\times 20=40$ meters

12. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $~20$ years. Four years ago, the product of their ages in years was $48$.

Let age of first friend = x years and let age of second friend $=\text{ }\left( 20-x \right)$ years

Four years ago, age of first friend $=\text{ }\left( x-4 \right)$ years

Four years ago, age of second friend $=\text{ }\left( 20-x \right)-4\text{ }=\text{ }\left( 16-x \right)$ years

According to given condition,

$\left( x-4 \right)\left( 16-x \right)=48$

$~\Rightarrow \text{ }16x-{{x}^{2}}\text{ }64+4x=48$

$\Rightarrow \text{ }20x-{{x}^{2}}\text{ }112=0$

$\Rightarrow \text{ }{{x}^{2}}-20x+112=0$

Comparing equation, ${{x}^{2}}-20x+112=0$ with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=-20\text{ }and\text{ }c=112$

Discriminant $={{b}^{2}}-4ac={{\left( -20 \right)}^{2}}\text{ }4\left( 1 \right)\left( 112 \right)=400\text{ }\text{ }448=-48<0$

The discriminant is less than zero which means we have no real roots for this equation.

Therefore, the given situation is not possible.

13. Value of$x$ for ${{x}^{2}}-8x+15=0$ is quadratic formula is

a). $3,\text{ }2$

b). $5,\text{ }2$

c). $5,\text{ }3$

d). $2,\text{ }3$

14. Discriminate of $\sqrt{3}{{x}^{2}}-2\sqrt{2}x-2\sqrt{3}=0$ is

a). $30$

b). $31$

c). $32$

d). $35$

15. Solve $12ab{{x}^{2}}-9{{a}^{2}}x+8{{b}^{2}}x-6ab=0$

$12ab{{x}^{2}}-9{{a}^{2}}x+8{{b}^{2}}x-6ab=0$

$\Rightarrow 3ax\left( 4bx-3x \right)+2b\left( 4bx-3x \right)=0$

$\Rightarrow \left( 4bx-3x \right)\left( 3ax+2b \right)=0$

$\Rightarrow 4bx-3a=0\,\,or\,\,3ax+2b=0$

$\therefore x=\dfrac{3a}{4b}\,or\,x=-\dfrac{2b}{3a}$

16. Solve for $x$ by quadratic formula${{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0$

${{p}^{2}}{{x}^{2}}+\left( {{p}^{2}}-{{q}^{2}} \right)x-{{q}^{2}}=0$

$a={{p}^{2}},\,b={{p}^{2}}-{{q}^{2}},\,c=-{{q}^{2}}$

$D={{b}^{2}}-4ac$

$=\left( {{p}^{2}}-{{q}^{2}} \right)-4\times {{p}^{2}}\left( -{{q}^{2}} \right)$

$={{p}^{4}}+{{q}^{2}}-2{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{2}}$

$={{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}$

$x=\dfrac{-b\pm \sqrt{D}}{29}$

$=\dfrac{-\left( {{p}^{2}}{{q}^{2}} \right)\pm \sqrt{{{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}}}{2\times {{p}^{2}}}$

$=\dfrac{-{{p}^{2}}+{{q}^{2}}+{{p}^{2}}+{{q}^{2}}}{2{{p}^{2}}}$

$or\,x=\dfrac{-{{p}^{2}}+{{q}^{2}}-{{p}^{2}}}{2{{p}^{2}}}$ $x=\dfrac{2{{q}^{2}}}{2{{p}^{2}}}\,\,or\,x=\dfrac{-2{{q}^{2}}}{2{{p}^{2}}}$

$x=\dfrac{{{q}^{2}}}{{{p}^{2}}}\,or\,x=-1$

17. Find the value of k for which the quadratic equation$k{{x}^{2}}+2x+1=0$ has real and

distinct root

$k{{x}^{2}}+2x+1=0$

$a=k,\,b=2,\,c=1$

$b={{b}^{2}}-4ac$

$={{\left( 2 \right)}^{2}}-4\times k\times 1=4-4k$

For real and distinct roots,

$D > 0$

$4-4k > 0$

$\Rightarrow -4k > -4$

$\therefore k < 1 $

18. If one root of the equations $2{{x}^{2}}+ax+3=0$ is $1$ , find the value of a.

a). $=-4$

b). $=-5$

c). $=-3$

d). $=-1$

19. Find k for which the quadratic equation$4{{x}^{2}}-3kx+1=0$ has equal root.

$=\pm \dfrac{3}{4}$

$=\dfrac{3}{4}$

$=\pm \dfrac{4}{3}$

$=\dfrac{2}{3}$

20. Determine the nature of the roots of the quadratic equation

$9{{a}^{2}}{{b}^{2}}{{x}^{2}}-24abcdx+16{{c}^{2}}{{d}^{2}}=0$

$={{\left( -24abcd \right)}^{2}}-4\times 9{{a}^{2}}{{b}^{2}}\times 16{{c}^{2}}{{d}^{2}}$

$=576{{a}^{2}}{{b}^{2}}{{c}^{2}}-376{{a}^{2}}{{b}^{2}}{{c}^{2}}{{d}^{2}}=0$

21. Find the discriminant of the equation $\left( x-1 \right)\left( 2x-1 \right)=0$

$\left( x-1 \right)\left( 2x-1 \right)=0$

$\Rightarrow 2{{x}^{2}}-x-2x+1=0$

$\Rightarrow 2{{x}^{2}}-3x+1=0$

$Here,a=2,\,b=-3,\,c=1$

$={{\left( -3 \right)}^{2}}-4\times 2\times 1$

$=9-8=1$

22. Find the value of k so that $\left( x-1 \right)$ is a factor of ${{k}^{2}}{{x}^{2}}-2kx-3$.

Let $P\left( x \right)={{k}^{2}}{{x}^{2}}-2kx-3$

$P\left( 1 \right)={{k}^{2}}{{\left( 1 \right)}^{2}}-2k\left( 1 \right)-3$

$\Rightarrow 0={{k}^{2}}-2k-3$

$\Rightarrow {{k}^{2}}-3k+k-3$

$\Rightarrow k\left( k-3 \right)+1\left( k-3 \right)=0$

$\Rightarrow \left( k-3 \right)\left( k+1 \right)=0$

$\therefore k=3\,or\,k=-1$

23. The product of two consecutive positive integers is $306$. Represent these in quadratic

a). ${{x}^{2}}+x-306=0$

b). ${{x}^{2}}-x-306=0$

c). ${{x}^{2}}+2x-106=0$

d). ${{x}^{2}}-x-106=0$

(a) ${{x}^{2}}+x-306=0$

24 . Which is a quadratic equation?

a). ${{x}^{2}}+x+2=0$

b). ${{x}^{3}}+{{x}^{2}}+2=0$

c). ${{x}^{4}}+{{x}^{2}}+2=0$

d). $x+2=0$

(a) ${{x}^{2}}+x+2=0$

25. The sum of two numbers is $16$. The sum of their reciprocals is $\dfrac{1}{3}$. Find the numbers.

Let no. be $x$

According to question,

$\dfrac{1}{x}+\dfrac{1}{16-x}=\dfrac{1}{3}$

$\Rightarrow \dfrac{16}{16x-{{x}^{2}}}=\dfrac{1}{3}$

$\Rightarrow {{x}^{2}}-16x+48=0$

$\Rightarrow {{x}^{2}}-12x-4x+48=0$

$\therefore x=12\,or\,x=4$

26. Solve for $x:\sqrt{217-x}=x-7$

$\sqrt{217-x}=\left( x-7 \right)$

$\Rightarrow 217-x={{x}^{2}}+49-14x$

$\Rightarrow {{x}^{2}}-14x+x+49-217=0$

$\Rightarrow {{x}^{2}}-13x-168=0$

$\Rightarrow {{x}^{2}}-21x+8x-168=0$

$\therefore x=21\,or\,x=-8$

27. Solve for x by factorization: $x+\dfrac{1}{x}=11\dfrac{1}{11}$

$\dfrac{{{x}^{2}}+1}{x}=\dfrac{122}{11}$

$\Rightarrow 11{{x}^{2}}-12x+11=0$

$\Rightarrow 11{{x}^{2}}-121x-1x+11=0$

$\Rightarrow 11x\left( x-11 \right)-1\left( x-11 \right)=0$

$\Rightarrow \left( 11x-1 \right)\left( x-11 \right)=0$

$\therefore x=11\,or\,x=\dfrac{1}{11}$

28. Find the ratio of the sum and product of the roots of $7{{x}^{2}}-12x+18=0$

Ans:

$7{{x}^{2}}-12x+18=0$

$\alpha +\beta =\dfrac{-b}{a}=\dfrac{12}{7}\,and\alpha \beta =\dfrac{c}{a}=\dfrac{18}{17}$

$\dfrac{\alpha +\beta }{\alpha \beta }=\dfrac{\dfrac{12}{7}}{\dfrac{18}{17}}=\dfrac{12}{7}\times \dfrac{17}{18}=\dfrac{34}{21}$

29. If $\alpha $ and$\beta $ are the roots of the equation ${{x}^{2}}+kx+12=0$, such that $\alpha -\beta =1$ , then

$\alpha +\beta =\dfrac{-k}{1},$

$\alpha -\beta =1$

$\alpha \beta =\dfrac{12}{1}$

${{\left( \alpha +\beta \right)}^{2}}={{\left( \alpha -\beta \right)}^{2}}+4\alpha \beta $

$\Rightarrow {{\left( -k \right)}^{2}}={{\left( 1 \right)}^{2}}+4\times 12$

$\Rightarrow {{k}^{2}}=49$

$k=\pm 7$

3 Marks Questions

1. Check whether the following are Quadratic Equations.

i). ${{\left( x+1 \right)}^{2}}=2\left( x-3 \right)$

${{\left( x+1 \right)}^{2}}=2\left( x-3 \right)\,\left\{ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\text{ } \right\}$

$\Rightarrow \text{ }{{x}^{2}}+1+2x=2x\text{ }6$

$\Rightarrow \text{ }{{x}^{2}}+7=0$

Here, degree of equation is $2$.

Therefore, it is a Quadratic Equation.

ii). ${{x}^{2}}-2x=\left( -2 \right)\left( 3-x \right)$

${{x}^{2}}-2x=\left( -2 \right)\left( 3-x \right)$

$\Rightarrow \text{ }{{x}^{2}}-2x=-6+2x$

$\Rightarrow \text{ }{{x}^{2}}-2x-2x+6=0$

$\Rightarrow \text{ }{{x}^{2}}-4x+6=0$

Here, degree of equation is $2.$

iii). $\left( x-2 \right)\left( x+1 \right)=\left( x-1 \right)\left( x+3 \right)$

(x−2)(x+1)=(x−1)(x+3)

$\Rightarrow \text{ }{{x}^{2}}+x-2x\text{ }2={{x}^{2}}+3x\text{ }x\text{ }3=0$

$\Rightarrow \text{ }{{x}^{2}}+x-2x\text{ }2-{{x}^{2}}-3x+x+3=0$

$\Rightarrow \text{ }x-2x\text{ }2-3x+x+3=0$

$\Rightarrow \text{ }-3x+1=0$

Here, degree of equation is $1.$

Therefore, it is not a Quadratic Equation.

iv). $~~\left( x-3 \right)\left( 2x+1 \right)=x\left( x+5 \right)$

$\left( x-3 \right)\left( 2x+1 \right)=x\left( x+5 \right)$

$\Rightarrow \text{ }2{{x}^{2}}+x-6x\text{ }3={{x}^{2}}+5x$

$\Rightarrow \text{ }2{{x}^{2}}+x-6x\text{ }3-{{x}^{2}}-5x=0$

$\Rightarrow \text{ }{{x}^{2}}-10x\text{ }3=0$

Here, degree of equation is 2.

Therefore, it is a quadratic equation.

v). $\left( 2x-1 \right)\left( x-3 \right)=\left( x+5 \right)\left( x-1 \right)$

$\left( 2x-1 \right)\left( x-3 \right)=\left( x+5 \right)\left( x-1 \right)\text{ }$

$\Rightarrow \text{ }2{{x}^{2}}-6x\text{ }x+3={{x}^{2}}\text{ }x+5x\text{ }5$

$\Rightarrow \text{ }{{x}^{2}}-11x+8=0$

Here, degree of Equation is$2$.

vi). ${{x}^{2}}+3x+1={{\left( x-2 \right)}^{2}}$

${{x}^{2}}+3x+1={{\left( x-2 \right)}^{2}}\,\,\,\,\,\text{ }\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \right\}$

$\Rightarrow \text{ }{{x}^{2}}\text{ }+3x+1={{x}^{2}}+4-4x$

$\Rightarrow \text{ }{{x}^{2}}+3x+1-{{x}^{2}}+4x\text{ }4=0$

$\Rightarrow \text{ }7x\text{ }3=0$

Here, degree of equation is$~1$.

vii). ${{\left( x+2 \right)}^{3}}=2x\left( {{x}^{2}}-1 \right)$

${{\left( x+2 \right)}^{3}}=2x\left( {{x}^{2}}-1 \right)\text{ }\,\,\,\,\,\,\,\left\{ {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \right\}$

$\Rightarrow \text{ }{{x}^{3}}\text{ }+{{2}^{3}}\text{ }+3\left( x \right)\left( 2 \right)\left( x+2 \right)=2x\left( {{x}^{2}}-1 \right)$

$\Rightarrow \text{ }{{x}^{3}}+8+6x\left( x+2 \right)=2{{x}^{3}}-2x$

$\Rightarrow \text{ }2{{x}^{3}}-2x-{{x}^{3}}\text{ }8-6{{x}^{2}}-12x=0$

$\Rightarrow \text{ }{{x}^{3}}-6{{x}^{2}}-14x\text{ }8=0$

Here, degree of Equation is 3.

Therefore, it is not a quadratic Equation.

viii). ${{x}^{3}}-4{{x}^{2}}\text{ }x+1={{\left( x-2 \right)}^{3}}\text{ }$

${{x}^{3}}-4{{x}^{2}}\text{ }x+1={{\left( x-2 \right)}^{3}}\,\,\,\,\text{ }\left\{ {{\left( a-b \right)}^{3}}\text{ }={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \right\}$

$\Rightarrow \text{ }{{x}^{3}}-4{{x}^{2}}\text{ }x+1={{x}^{3}}-{{2}^{3}}\text{ }3\left( x \right)\left( 2 \right)\left( x-2 \right)$

$\Rightarrow \text{ }-4{{x}^{2}}\text{ }x+1=-8-6{{x}^{2}}+12x$

$\Rightarrow \text{ }2{{x}^{2}}-13x+9=0$

Here, degree of Equation is $2$.

2. Represent the following situations in the form of Quadratic Equations:

i). The area of the rectangular plot is 528 $m^2$. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

We are given that area of a rectangular plot is $528{{m}^{2}}$

Let the breadth of the rectangular plot be$~x$ meters

Length is one more than twice its breadth

Therefore, length of rectangular plot is $\left( 2x+1 \right)$meters

Area of rectangle$=$length $\times $ breadth

$\Rightarrow \text{ }528=x\left( 2x+1 \right)$

$\Rightarrow \text{ }528=2{{x}^{2}}+x$

$\Rightarrow \text{ }2{{x}^{2}}+x\text{ }528=0$

This is a Quadratic Equation.

ii). The product of two consecutive numbers is 306. We need to find the integers.

Let two consecutive numbers be $x\,\text{and}\,\left( x+1 \right).$

It is given that x(x+1) = 306

$\Rightarrow \text{ }{{x}^{2}}+x=306$

$\Rightarrow \text{ }{{x}^{2}}+x\text{ }306=0$

iii). Rohan's mother is 26 years older than him. The product of their ages (in years) after 3 years will be 360. We would like to find Rohan's present age.

Let present age of Rohan $=\text{ }x$years

Let present age of Rohan's mother $=\text{ }\left( x\text{ }+26 \right)$ years

Age of Rohan after$~3$years $=\text{ }\left( x+3 \right)$ years

Age of Rohan's mother after $~3$ years $=\text{ }x+26+3\text{ }=\text{ }\left( x+29 \right)$ years

According to given condition:

$\left( x+3 \right)\left( x+29 \right)=360$

$\Rightarrow \text{ }{{x}^{2}}+29x+3x+87=360$

$\Rightarrow \text{ }{{x}^{2}}+32x\text{ }273=0$

iv). A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Let the speed of the train be $x$ km/h

Time taken by train to cover 480 km $=\text{ }480x$hours

If, speed had been 8km/h less than time taken would be $\left( 480x-8 \right)$ hours. According to given condition, if speed had been $8$km/h less than time taken is $3$hours less.

Therefore, $480x\text{ }8=480x+3$

$\Rightarrow \text{ }480\left( 1x\text{ }8-1x \right)=3$

$\Rightarrow \text{ }480\left( x\text{ }x+8 \right)\text{ }\left( x \right)\text{ }\left( x-8 \right)=3$

$\Rightarrow \text{ }480\times 8=3\left( x \right)\left( x-8 \right)$

$~\Rightarrow \text{ }3840=3{{x}^{2}}-24x$

Dividing equation by$3$, we get

$\Rightarrow \text{ }{{x}^{2}}-8x\text{ }1280=0$

This is a Quadratic Equation.

3. Find the roots of the following Quadratic Equations by factorization.

i). ${{x}^{2}}-3x\text{ }10=0$

${{x}^{2}}-3x\text{ }10=0$

$\Rightarrow \text{ }{{x}^{2}}-5x+2x\text{ }10=0$

$\Rightarrow \text{ }x\left( x-5 \right)+2\left( x-5 \right)=0$

$\Rightarrow \text{ }\left( x-5 \right)\left( x+2 \right)=0$

$\Rightarrow \text{ }x=5,-2$

ii). $2{{x}^{2}}+x\text{ }6=0$

$2{{x}^{2}}+x\text{ }6=0$

$\Rightarrow \text{ }2{{x}^{2}}+4x-3x\text{ }6=0$

$\Rightarrow \text{ }2x\left( x+2 \right)\text{ }\text{ }3\left( x+2 \right)=0$

$\Rightarrow \text{ }\left( 2x-3 \right)\left( x+2 \right)=0$

$\Rightarrow x=\dfrac{3}{2},2$

iii). $\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$

$\sqrt{2}{{x}^{2}}+7x+5\sqrt{2}=0$

$\Rightarrow \sqrt{2}{{x}^{2}}+2x+5x+5\sqrt{2}=0$

$\Rightarrow \sqrt{2}{{x}^{2}}\left( x+\sqrt{2} \right)+5\left( x+\sqrt{2} \right)=0$

$\Rightarrow \left( \sqrt{2}x+5 \right)\left( x+\sqrt{2} \right)=0$

$\Rightarrow x=\dfrac{-5}{\sqrt{2}},-\sqrt{2}$

$\Rightarrow x=\dfrac{-5}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}},-\sqrt{2}$

$\Rightarrow x=\dfrac{-5\sqrt{2}}{2},-\sqrt{2}$

iv). $2{{x}^{2}}-x+\dfrac{1}{8}=0$

$2{{x}^{2}}-x+\dfrac{1}{8}=0$

$\Rightarrow \dfrac{16{{x}^{2}}-8x+1}{8}=0$

$\Rightarrow \text{ }16{{x}^{2}}-8x+1=0$

$\Rightarrow \text{ }16{{x}^{2}}-4x-4x+1=0$

$\Rightarrow \text{ }4x\left( 4x-1 \right)\text{ }\text{ }1\left( 4x-1 \right)=0$

$\Rightarrow \text{ }\left( 4x-1 \right)\left( 4x-1 \right)=0$

$\Rightarrow \text{ }x=\text{ }\dfrac{1}{4},\dfrac{1}{4}$

v). $100{{x}^{2}}-20x+1=0$

$100{{x}^{2}}-20x+1=0$

$\Rightarrow \text{ }100{{x}^{2}}-10x-10x+1=0$

$\Rightarrow \text{ }10x\left( 10x-1 \right)\text{ }\text{ }1\left( 10x-1 \right)=0$

$\Rightarrow \text{ }\left( 10x-1 \right)\left( 10x-1 \right)=0$

$\therefore \,x=\dfrac{1}{10},\dfrac{1}{10}$

4. Find the roots of the following equations:

i). $\dfrac{x-1}{x}=3,x\ne 0$

$x-\dfrac{1}{x}=3\,\,where\,x\ne 0$

$\Rightarrow \dfrac{{{x}^{2}}-1}{x}=3$

$\Rightarrow \text{ }x{{\text{ }}^{2}}\text{ }1=3x$

$\Rightarrow \text{ }{{x}^{2}}-3x\text{ }1=0$

Comparing equation ${{x}^{2}}-3x\text{ }1=0$with general form$a{{x}^{2}}+bx+c=0$,

We get $a=1,b=-3\text{ }and\text{ }c=-1$

Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,

$x=\dfrac{3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{3\pm \sqrt{13}}{2}$

$\Rightarrow x=\dfrac{3+\sqrt{13}}{2},\dfrac{3-\sqrt{13}}{2}$

(ii). $\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30},x\ne -4,7$

$\dfrac{1}{x+4}-\dfrac{1}{x-7}=\dfrac{11}{30}\,where\,x\ne -4,7$

$\Rightarrow \text{ }-30={{x}^{2}}-7x+4x28$

$\Rightarrow \text{ }{{x}^{2}}-3x+2=0$

Comparing equation$~{{x}^{2}}-3x+2=0$ with general form$a{{x}^{2}}+bx+c=0$,

We get $a=1,b=-3\text{ }and\text{ }c=2$

Using quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,

$x=\dfrac{3\pm \sqrt{{{\left( 3 \right)}^{2}}-4\left( 1 \right)\left( 2 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{3\pm \sqrt{1}}{2}$

$\Rightarrow x=\dfrac{3+\sqrt{1}}{2},\dfrac{3-\sqrt{1}}{2}$

$\Rightarrow x=2,1$

5. The sum of reciprocals of Rehman's ages (in years) $3$ years ago and $5$ years from now is $13$. Find his present age.

Let present age of Rehman$=\text{ }x$ years

Age of Rehman $3$ years ago $=\text{ }\left( x-3 \right)$ years.

Age of Rehman after $5$ years $=\text{ }\left( x+5 \right)$ years

According to the given condition:

$\dfrac{1}{x-3}+\dfrac{1}{x+5}=\dfrac{1}{3}$

$\Rightarrow \dfrac{\left( x+5 \right)+\left( x-3 \right)}{\left( x-3 \right)\left( x+5 \right)}=\dfrac{1}{3}$

$\Rightarrow \text{ }3\left( 2x+2 \right)\text{ }=\left( x-3 \right)\left( x+5 \right)\text{ }$

$\Rightarrow \text{ }6x+6={{x}^{2}}-3x+5x-15$

$\Rightarrow \text{ }{{x}^{2}}-4x\text{ }15\text{ }\text{ }6=0$

$\Rightarrow {{x}^{2}}-4x\text{ }21=0$

Comparing quadratic equation x ${{x}^{2}}-4x\text{ }21=0$ with general form $a{{x}^{2}}+bx+c=0$, We get $a=1,b=-4\text{ }and\text{ }c=-21$

Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( -21 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{4\pm \sqrt{16+84}}{2}$

$\Rightarrow x=\dfrac{4\pm \sqrt{100}}{2}=\dfrac{4\pm 10}{2}$

$\Rightarrow x=\dfrac{4+10}{2},\dfrac{4-10}{2}$

$\therefore x=7,-3$

We discard$x=-3$ .Since age cannot be in negative.

Therefore, present age of Rehman is $7$ years

6. Two water taps together can fill a tank in $9\dfrac{3}{8}$hours. The tap of larger diameter takes $10$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Let time taken by tap of smaller diameter to fill the tank $=\text{ }x$ hours

Let time taken by tap of larger diameter to fill the tank $=\text{ }\left( x\text{ }10 \right)$ hours

It means that tap of smaller diameter fills ${{\dfrac{1}{x}}^{th}}$ part of tank in $1$ hour. … (1)

And, tap of larger diameter fills ${{\dfrac{1}{x-10}}^{th}}$ part of tank in $1$ hour. … (2)

When two taps are used together, they fill tank in$~758$ hours

In 1 hour, they fill${{\dfrac{8}{75}}^{th}}$ part of tank $\left[ \dfrac{1}{\dfrac{75}{8}}=\dfrac{8}{75} \right]$ … (3)

From (1), (2) and (3),

$\dfrac{1}{x}+\dfrac{1}{x-10}=\dfrac{8}{75}$

$\Rightarrow \dfrac{x-10+x}{x\left( x-10 \right)}=\dfrac{8}{75}$

$\Rightarrow \text{ }75\left( 2x-10 \right)=8\left( {{x}^{2}}-10x \right)$

$\Rightarrow \text{ }150x\text{ }750=8{{x}^{2}}-80x$

$\Rightarrow \text{ }8{{x}^{2}}\text{ }-80x-150x+750=0$

Comparing equation $4{{x}^{2}}\text{ }-115x+375=0$ with general equation$a{{x}^{2}}+bx+c=0$,

We get $a=4,b=-115\,and\,c=375$

Applying quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{115\pm \sqrt{{{\left( -115 \right)}^{2}}-4\left( 4 \right)\left( 375 \right)}}{2\times 4}$

$\Rightarrow x=\dfrac{115\pm \sqrt{13225-6000}}{8}$

$\Rightarrow \dfrac{115\pm \sqrt{7225}}{8}$

$\Rightarrow \dfrac{115+85}{8},\dfrac{115-85}{8}$

$\therefore x=25,3.75$

Time taken by larger tap $=\text{ }x\text{ }10=3.75\text{ }\text{ }10=-6.25$ hours

Time cannot be in negative. Therefore, we ignore this value.

Time taken by larger tap $=\text{ }x\text{ }10=25\text{ }\text{ }10=15$ hours

Therefore, time taken by larger tap is $15$ hours and time taken by smaller tap is $25$ hours.

7. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.

i). $2{{x}^{2}}\text{ }3x\text{ }+\text{ }5\text{ }=\text{ }0$

$2{{x}^{2}}\text{ }3x\text{ }+\text{ }5\text{ }=\text{ }0$

Comparing this equation with general equation$a{{x}^{2}}+bx+c=0$ ,

We get $a=2,b=-3\text{ }and\text{ }c=5$

Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -3 \right)}^{2}}\text{ }4\left( 2 \right)\left( 5 \right)=9\text{ }\text{ }40=-31$

Discriminant is less than 0 which means the equation has no real roots.

ii). $3{{x}^{2}}-4\sqrt{3}x+4=0$

$3{{x}^{2}}-4\sqrt{3}x+4=0$

Comparing this equation with general equation $a{{x}^{2}}+bx+c=0$,

We get $a=3,b=-4\sqrt{3}\text{ }and\text{ }c=4$

Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -4\sqrt{3} \right)}^{2}}-4\left( 3 \right)\left( 4 \right)=48\text{ }\text{ }48=0$

Discriminant is equal to zero which means equations have equal real roots. Applying quadratic$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find roots,

$x=\dfrac{4\sqrt{3}\pm \sqrt{0}}{6}=\dfrac{2\sqrt{3}}{3}$

Because, equation has two equal roots, it means $x=\dfrac{2\sqrt{3}}{3},\dfrac{2\sqrt{3}}{3}$

iii). $2{{x}^{2}}\text{ }+6x\text{ }+\text{ }3\text{ }=\text{ }0$

$2{{x}^{2}}\text{ }+6x\text{ }+\text{ }3\text{ }=\text{ }0$

Comparing equation with general equation$a{{x}^{2}}+bx+c=0$ ,

We get $a=2,b=-6,\text{ }and\text{ }c=3$

Discriminant $=\text{ }{{b}^{2}}-4ac={{\left( -6 \right)}^{2}} - \text{ }4\left( 2 \right)\left( 3 \right)=36\text{ }-\text{ }24=12$

Value of the discriminant is greater than zero.

Therefore, the equation has distinct and real roots.

Applying quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find roots,

$x=\dfrac{6\pm \sqrt{12}}{4}=\dfrac{6\pm 2\sqrt{3}}{4}$

$\Rightarrow x=\dfrac{3\pm \sqrt{3}}{2}$

$\Rightarrow x=\dfrac{3+\sqrt{3}}{2},\dfrac{3-\sqrt{3}}{2}$

8. If $-4$ is a root of the quadratic equation and the quadratic equation${{x}^{2}}+px-4$ has equal root, find the value of $k.$

$-4$is root of ${{x}^{2}}+px-4=0$

$\therefore {{\left( -4 \right)}^{2}}+p\left( -4 \right)-4=0$

$\Rightarrow 16-4p-4=0$

$\Rightarrow -4p=-12$

$\Rightarrow p=3$

${{x}^{2}}+px+k=0$ (Given)

${{x}^{2}}+3x+k=0$

$\Rightarrow 0={{\left( 3 \right)}^{2}}-4\times 1\times k$ [For equal roots D = 0]

$\Rightarrow 4k=9$

$\Rightarrow k=\dfrac{9}{4}$

9. Solve for $x:{{5}^{x+1}}+{{5}^{1-x}}=26$

${{5}^{x+1}}+{{5}^{1-x}}=26$

${{5}^{x}}{{.5}^{1}}+{{5}^{1}}{{.5}^{-x}}=26$

$\Rightarrow {{5}^{x}}.5+\dfrac{{{5}^{1}}}{{{5}^{x}}}=26$

Put ${{5}^{x}}=y$

$\dfrac{5y}{1}+\dfrac{5}{y}=26$

$\Rightarrow 5{{y}^{2}}-26y+5=0$

$\Rightarrow 5{{y}^{2}}-25y-y+5=0$

$\Rightarrow 5y\left( y-5 \right)-1\left( y-5 \right)=0$

$\Rightarrow $ $\left( y-5 \right)\left( 5y-1 \right)=0$

$\Rightarrow y=5$ or $y=\dfrac{1}{5}$

But

${{5}^{x}}={{5}^{1}}$ and ${{5}^{x}}=\dfrac{1}{5}$

$\Rightarrow x=1$ and ${{5}^{x}}={{5}^{-1}}\Rightarrow x=-1$

10. $\dfrac{1}{p+q+x}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{x}$ solve for$x$ by factorization method.

$\dfrac{1}{p+q+x}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{x}$

$\Rightarrow \dfrac{1}{p+q+x}-\dfrac{1}{x}=\dfrac{1}{p}+\dfrac{1}{q}$

$\Rightarrow \dfrac{x-p-q-x}{{{x}^{2}}+px+qx}=\dfrac{p+q}{pq}$

$\Rightarrow \dfrac{-\left( p+q \right)}{{{x}^{2}}+px+qx}=\dfrac{p+q}{pq}$

$\Rightarrow \dfrac{-1}{{{x}^{2}}+px+qx}=\dfrac{1}{pq}$

$\Rightarrow {{x}^{2}}+px+qx=-pq$

$\Rightarrow {{x}^{2}}+px+qx+pq=0$

$\Rightarrow x\left( x+p \right)+q\left( x+p \right)=0$

$\Rightarrow \left( x+p \right)\left( x+q \right)=0$

$\therefore x=-p\,or\,x=-q$

11. $5{{x}^{2}}-6x-2=0$, solve for$x$ by the method of completing the square.

$5{{x}^{2}}-6x-2=0$

$\Rightarrow {{x}^{2}}-\dfrac{6}{5}x-\dfrac{2}{5}=0$

$\Rightarrow {{x}^{2}}-\dfrac{6}{5}x+{{\left( \dfrac{3}{5} \right)}^{2}}-\dfrac{2}{5}=0$

$\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9}{25}+\dfrac{2}{5}$

$\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{9+10}{25}$

$\Rightarrow {{\left( x-\dfrac{3}{5} \right)}^{2}}=\dfrac{19}{25}$

$\Rightarrow x-\dfrac{3}{5}=\pm \dfrac{\sqrt{19}}{5}$

$\Rightarrow x=\dfrac{3}{5}\pm \dfrac{\sqrt{19}}{5}$

$\Rightarrow x=\dfrac{3+\sqrt{19}}{5}\,or\,x=\dfrac{3-\sqrt{19}}{5}$

12. Solve for$x:{{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$

${{a}^{2}}{{b}^{2}}{{x}^{2}}+{{b}^{2}}x-{{a}^{2}}x-1=0$

$\Rightarrow {{b}^{2}}x\left( {{a}^{2}}x+1 \right)-1\left( {{a}^{2}}x-1 \right)=0$

$\Rightarrow \left( {{a}^{2}}x+1 \right)\left( {{b}^{2}}x-1 \right)=0$

$\therefore x=\dfrac{-1}{{{a}^{2}}}\,or\,x=\dfrac{-1}{{{b}^{2}}}$

13. Using quadratic formula, solve for $x:9{{x}^{2}}-9\left( a+b \right)x+\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)=0$

$={{\left( -9\left( a+b \right) \right)}^{2}}-4\times 9\times \left( 2{{a}^{2}}+5ab+2a{{b}^{2}} \right)$

$=81{{\left( a+b \right)}^{2}}-36\left( 2{{a}^{2}}+5ab+2{{b}^{2}} \right)$

$=9\left[ 9\left( {{a}^{2}}+{{b}^{2}}+2ab-8{{a}^{2}}-20ab-8{{b}^{2}} \right) \right]$

$=9\left[ {{a}^{2}}+{{b}^{2}}-2ab \right]$

$=9{{\left( a-b \right)}^{2}}$

$x=\dfrac{-b\pm \sqrt{D}}{2a}=\dfrac{9\left( a+b \right)\pm \sqrt{9{{\left( a-b \right)}^{2}}}}{2\times 9}$

$\Rightarrow x=3\dfrac{\left[ 3\left( a+b \right)\pm \left( a-b \right) \right]}{2\times 9}$

$\Rightarrow x=\dfrac{\left( 3a+3b \right)\pm \left( a-b \right)}{6}$

$\Rightarrow x=\dfrac{3a+3b+a-b}{6}\,or\,x=\dfrac{3a+3b+a-b}{6}$

$\Rightarrow x=\dfrac{4a+2b}{6}\,or\,x=\dfrac{4a+2b}{6}$

$\Rightarrow x=\dfrac{2a+b}{3}\,or\,x=\dfrac{2a+b}{3}$

14. In a cricket match, Kapil took one wicket less than twice the number of wickets taken by Ravi. If the product of the numbers of wickets taken by these two is 15, find the number of wickets taken by each.

Let no. of wicket taken by Ravi $=x$

of wicket taken by Kapil $=2x-1$

$\left( 2x-1 \right)x=15$

$\Rightarrow 2{{x}^{2}}-x-15=0$

$\therefore x=3\,or\,x=\dfrac{-5}{2}$ (Neglects)

So, no. of wickets taken by Ravi is $x=3$

15 . The sum of a number and its reciprocal is $\dfrac{17}{4}$ . Find the number.

$\dfrac{x}{1}+\dfrac{1}{x}=\dfrac{17}{4}$

$\Rightarrow \dfrac{{{x}^{2}}+1}{x}=\dfrac{17}{4}$

$\Rightarrow 4{{x}^{2}}+4=17x$

$\Rightarrow 4{{x}^{2}}-17x+4=0$

$\Rightarrow 4{{x}^{2}}-16x-x+4=0$

$\Rightarrow 4x\left( x-4 \right)-1\left( x-4 \right)=0$

$\Rightarrow \left( x-4 \right)\left( 4x-1 \right)=0$

$\therefore x=4\,or\,x=\dfrac{1}{4}$

4 Marks Questions

1. Find the roots of the following Quadratic Equations by applying quadratic formulas.

i). $2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$

$2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$

Comparing quadratic equation $2{{x}^{2}}-\text{ }7x+\text{ }3\text{ }=\text{ }0$with general form $a{{x}^{2}}+bx+c=0$, we get $a=2,\text{ }b=-7\text{ }and\text{ }c=3$

Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{7\pm \sqrt{{{\left( -7 \right)}^{2}}-4\left( 2 \right)\left( 3 \right)}}{2\times 2}$

$\Rightarrow x=\dfrac{7\pm \sqrt{49-24}}{4}$

$\Rightarrow x=\dfrac{7\pm 5}{4}$

$\Rightarrow x=\dfrac{7+5}{4},\dfrac{7-5}{4}$

$\therefore x=3,\dfrac{1}{2}$

ii). $2{{x}^{2}}+\text{ }x\text{ }\text{ }4\text{ }=\text{ }0$

Comparing quadratic equation $2{{x}^{2}}+\text{ }x\text{ }\text{ }4\text{ }=\text{ }0$with the general form$a{{x}^{2}}+bx+c=0$, we get $a=2,\text{ }b=1\text{ }and\text{ }c=-4$

Putting these values in quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$ x=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\left( 2 \right)\left( -4 \right)}}{2\times 2} $

$ \Rightarrow x=\dfrac{-1\pm \sqrt{33}}{4} $

$ \Rightarrow x=\dfrac{-1-\sqrt{33}}{4},\dfrac{-1+\sqrt{33}}{4} $

iii). $4{{x}^{2}}+4\sqrt{3}x+3=0$

Comparing quadratic equation $4{{x}^{2}}+4\sqrt{3}x+3=0$ with the general form$a{{x}^{2}}+bx+c=0$, we get $a=4,\text{ }b=4\sqrt{3}\text{ }and\text{ }c=3$

$x=\dfrac{-4\sqrt{3}\pm \sqrt{{{\left( 4\sqrt{3} \right)}^{2}}-4\left( 4 \right)\left( 3 \right)}}{2\times 4}$

$\Rightarrow x=\dfrac{-4\sqrt{3}\pm \sqrt{0}}{8}$

$\Rightarrow x=\dfrac{-\sqrt{3}}{2}$

A quadratic equation has two roots. Here, both the roots are equal.

Therefore,$x=\dfrac{-\sqrt{3}}{2},\dfrac{-\sqrt{3}}{2}$

iv). $2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$

$2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$

Comparing quadratic equation $2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$ with the general form$a{{x}^{2}}+bx+c=0$, we get $a=2,b=1\text{ }and\text{ }c=\text{ }4$

$x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 2 \right)\left( 4 \right)}}{2\times 2}$

$\Rightarrow x=\dfrac{-1\pm \sqrt{-31}}{4}$

But, the square root of a negative numbers is not defined.

Therefore, Quadratic Equation $2{{x}^{2}}+\text{ }x\text{ }+\text{ }4\text{ }=\text{ }0$ has no solution.

2. An express train takes$~1$ hour less than a passenger train to travel $132$ km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If, the average speed of the express train is $11$ km/h more than that of the passenger train, find the average speed of two trains

Ans: Let average speed of passenger train $=\text{ }x$ km/h

Let average speed of express train $=\text{ }\left( x+11 \right)$ km/h

Time taken by passenger train to cover $132$ km $=$ $\dfrac{132}{x}$ hours

Time taken by express train to cover$132$km $=\left( \dfrac{132}{x+11} \right)$ hours

According to the given condition

$\dfrac{132}{x}=\dfrac{132}{x+11}+1$

$\Rightarrow 132\left( \dfrac{1}{x}-\dfrac{1}{x+11} \right)=1$

$132\left( \dfrac{x+11-x}{x\left( +11 \right)} \right)=1$

$\Rightarrow \text{ }132\left( 11 \right)=x\left( x+11 \right)$

$\Rightarrow \text{ }1452={{x}^{2}}+11x$

$\Rightarrow \text{ }{{x}^{2}}+11x\text{ }1452=0$

Comparing equation ${{x}^{2}}+11x\text{ }1452=0$with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=11\text{ }and\text{ }c=-1452$

Applying Quadratic Formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\dfrac{-11\pm \sqrt{{{\left( 11 \right)}^{2}}-4\left( 1 \right)\left( -1452 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{-11\pm \sqrt{121+5808}}{2}$

$\Rightarrow x=\dfrac{-11\pm \sqrt{5929}}{2}$

$\Rightarrow x=\dfrac{-11\pm 77}{2}$

$\Rightarrow x=\dfrac{-11+77}{2},\dfrac{-11-77}{2}$

$\therefore x=33,-44$

As speed cannot be negative. Therefore, speed of passenger train $=\text{ }33$ km/h

And, speed of express train $=\text{ }x+11=33+11=44$ km/h

3. Sum of areas of two squares is $468\text{ }{{m}^{2}}$. If, the difference of their perimeters is $24$ meters, find the sides of the two squares.

Let perimeter of first square $=\text{ }x$ meters

Let perimeter of second square $=\text{ }\left( x+24 \right)$ meters

Length of side of first square$~=$$\dfrac{x}{4}$ meters {Perimeter of square $=\text{ }4\text{ }\times $ length of side}

Length of side of second square $=$ $=\left( \dfrac{x+24}{4} \right)$ meters

Area of first square $= side \times side $

=$\dfrac{x}{4}\times \dfrac{x}{4}=\dfrac{{{x}^{2}}}{16}{{m}^{2}}$

Area of second square $={{\left( \dfrac{x+24}{4} \right)}^{2}}{{m}^{2}}$

$\dfrac{{{x}^{2}}}{16}+{{\left( \dfrac{x+24}{4} \right)}^{2}}=468$

$\Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{x}^{2}}+576+48x}{16}=468$

$\Rightarrow \dfrac{{{x}^{2}}+{{x}^{2}}+576+48x}{16}=468$

$\Rightarrow \text{ }2{{x}^{2}}+576+48x=468\times 16$

$\Rightarrow \text{ }2{{x}^{2}}+48x+576=7488\text{ }$

$\Rightarrow \text{ }2{{x}^{2}}+48x\text{ }6912=0\text{ }$

$\Rightarrow \text{ }{{x}^{2}}+24x\text{ }3456=0$

Comparing equation${{x}^{2}}+24x\text{ }3456=0$ with standard form$a{{x}^{2}}+bx+c=0$, We get $a=1,b=24\text{ }and\,c=\text{ }-3456$

$x=\dfrac{-24\pm \sqrt{{{\left( 24 \right)}^{2}}-4\left( 1 \right)\left( -3456 \right)}}{2\times 1}$

$\Rightarrow x=\dfrac{-24\pm \sqrt{576+13824}}{2}$

$\Rightarrow x=\dfrac{-24\pm \sqrt{14400}}{2}=\dfrac{-24\pm 120}{2}$

$\Rightarrow x=\dfrac{-24+120}{2},\dfrac{-24-120}{2}$

$\therefore x=48,-72$

Perimeter of square cannot be in negative. Therefore, we discard$x=-72$.

Therefore, perimeter of first square $=\text{ }48$meters

And, Perimeter of second square $=\text{ }x+24=48+24=72$meters

$\Rightarrow $ Side of First square $=\dfrac{Perimeter}{4}=\dfrac{48}{4}=12m$

And, Side of second Square $=\dfrac{Perimeter}{4}=\dfrac{72}{4}=18m$

4. Is it possible to design a rectangular park of perimeter $80$ meters and area$400\text{ }{{m}^{2}}$. If so, find its length and breadth.

Let length of park $=\text{ }x$ meters

We are given area of rectangular park $=\text{ }400\text{ }{{m}^{2}}$

Therefore, breadth of park $=$ $\dfrac{400}{x}$ meters {Area of rectangle$~=$ length $\times $ breadth}

Perimeter of rectangular park $=\text{ }2$ (length$+$breath)$=$ $\left( x+\dfrac{400}{x} \right)$ meters

We are given perimeter of rectangle $=\text{ }80$ meters

According to condition:

$2\left( x+\dfrac{400}{x} \right)=80$

$\Rightarrow 2\left( \dfrac{{{x}^{2}}+400}{x} \right)=80$

$\Rightarrow \text{ }2{{x}^{2}}+800=80x$

$\Rightarrow \text{ }2{{x}^{2}}-80x+800=0$

$\Rightarrow \text{ }{{x}^{2}}-40x+400=0$

Comparing equation, ${{x}^{2}}-40x+400=0$ with general quadratic equation$a{{x}^{2}}+bx+c=0$, we get $a=1,b=-40\text{ }and\text{ }c=400$

Discriminant$=\text{ }{{b}^{2}}-4ac={{\left( -40 \right)}^{2}}\text{ }4\left( 1 \right)\left( 400 \right)=1600\text{ }\text{ }1600=0$

Discriminant is equal to$0$ .

Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter $80$meters and area$400\text{ }{{m}^{2}}$.

Using quadratic formula$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve equation,s

$x=\dfrac{40\pm \sqrt{0}}{2}=\dfrac{42}{2}=20$

Here, both the roots are equal to$20$.

Therefore, length of rectangular park $=\text{ }20$meters

Breadth of rectangular park$=\dfrac{400}{x}=\dfrac{400}{20}=20m$

5. If I had walked $1$ km per hour faster, I would have taken $10$ minutes less to walk $2$ km. Find the rate of my walking.

Ans: Distance $=\text{ }2$ km

Let speed $=x$ km/hr.

New speed$~=\text{ }\left( x+1 \right)$ km/hr.

Time taken by normal speed $=\dfrac{2}{x}\,hr$

Time taken by new speed = $\dfrac{2}{x+1}hr$

$\dfrac{2}{x}-\dfrac{2}{x+1}=\dfrac{10}{60}$

$\Rightarrow \dfrac{2x+2-2x}{{{x}^{2}}+x}=\dfrac{1}{6}$

$\Rightarrow {{x}^{2}}+x=12$

$\Rightarrow {{x}^{2}}+x-12=0$

$\Rightarrow {{x}^{2}}+4x-3x-12=0$

$\Rightarrow x\left( x+4 \right)-3\left( x+4 \right)=0$

$\Rightarrow \left( x+4 \right)\left( x-3 \right)=0$

$\therefore x=-4\,or\,x=3$

So, speed is $x=3$ km/hr

6. A takes $6$ days less than the time taken by B to finish a piece of work. If both A and B together can finish it in $4$ days, find the time taken by B to finish the work.

Ans: Let B takes $x$ days to finish the work, then A alone can finish it in $\left( x-6 \right)$ days

$\dfrac{1}{x}+\dfrac{1}{x-6}=\dfrac{1}{4}$

$\Rightarrow \dfrac{x-6+x}{{{x}^{2}}-6x}=\dfrac{1}{4}$

$\Rightarrow \dfrac{2x-6}{{{x}^{2}}-6x}=\dfrac{1}{4}$

$\Rightarrow {{x}^{2}}-6x=8x-24$

$\Rightarrow {{x}^{2}}-14x+24=0$

$\Rightarrow {{x}^{2}}-12x-2x+24=0$

$=x\left( x-12 \right)-2\left( x-12 \right)=0$

$\Rightarrow \left( x-12 \right)\left( x-2 \right)=0$

$\therefore x=12\,or\,x=2$

$x=2$ (Neglect)

So, B takes$x=12$ days.

7. A plane left $30$ minutes later than the schedule time and in order to reach its $59$ destination $1500$ km away in time it has to increase its speed by $250$km/hr from its usual speed. Find its usual speed.

Ans: Let usual speed$=x$ km/hr

New speed $=\left( x+250 \right)$ km/hr

Total distance $=\text{ }1500$ km

Time taken by usual speed $=\dfrac{1500}{x}$ hr

Time taken by new speed $=\dfrac{1500}{x+250}$ hr

$\dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{1}{2}$

$\Rightarrow \dfrac{1500x+1500\times 250-1500x}{{{x}^{2}}+250x}=\dfrac{1}{2}$

$\Rightarrow {{x}^{2}}+250x=\dfrac{1500\times 250}{2}$

$\Rightarrow {{x}^{2}}+250x=750000$

$\Rightarrow {{x}^{2}}+250x-750000=0$

$\Rightarrow x^{2}+1000x-750x-750000=0$

$\Rightarrow x\left( x+1000 \right)-750\left( x+1000 \right)=0$

$\Rightarrow x=750\,\,or\,x=-1000$

Therefore, usual speed is $750$ km/hr, $-1000$is neglected as speed cannot be negative.

8. A motor boat, whose speed is $15$ km/hr in still water, goes $30$ km downstream and comes back in a total time of $4$ hr $30$ minutes, find the speed of the stream.

Speed of motor boat in still water $=\text{ }15$ km/hr

Speed of stream $=x$ km/hr

Speed in downward direction $15+x$

Speed in downward direction $15-x$

According to question,

$\dfrac{30}{15+x}+\dfrac{30}{15-x}=4\dfrac{1}{2}$

$\Rightarrow \dfrac{30\left( 15-x \right)+30\left( 15+x \right)}{\left( 15+x \right)\left( 15-x \right)}=\dfrac{9}{2}$

$\Rightarrow \dfrac{450-30x+450+30x}{225-{{x}^{2}}}=\dfrac{9}{2}$

$\Rightarrow 9\left( 225-{{x}^{2}} \right)=1800$

$\Rightarrow 225-{{x}^{2}}=200$

$\therefore x=5$

Speed of stream$~=\text{ }5$ km/hr.

9. A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously fill the pool in the same time during which the pool is the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.

Let $x$ be the number of hours required by the second pipe alone to till the pool and first pipe $\left( x+5 \right)$ hour while third pipe$\left( x-4 \right)$ hour

$\dfrac{1}{x+5}+\dfrac{1}{x}=\dfrac{1}{x-4}$

$\Rightarrow \dfrac{x+x+5}{{{x}^{2}}+5x}=\dfrac{1}{x-4}$

$\Rightarrow {{x}^{2}}-8x-20=0$

$\Rightarrow {{x}^{2}}-10x+2x-20=0$

$\Rightarrow x\left( x-10 \right)+2\left( x-10 \right)=0$

$\Rightarrow \left( x-10 \right)\left( x+2 \right)=0$

$\Rightarrow x=10\,or\,x=-2\left( Neglected \right)$

10. A two-digit number is such that the product of its digits is$18$. When $63$ is subtracted from the number the digit interchanges their places. Find the number

Let digit on unit’s place $=x$

Digit on ten’s place $=y$

$xy=18$ (given)

Number = 10.$y+x$

$=10\left( \dfrac{18}{x} \right)+x$

$10\left( \dfrac{18}{x} \right)+x-63=10x+\dfrac{18}{x}$

$\Rightarrow \dfrac{180}{x}+\dfrac{x-63}{1}=\dfrac{10{{x}^{2}}+18}{x}$

$\Rightarrow \dfrac{180+{{x}^{2}}-63x}{x}=\dfrac{10{{x}^{2}}+18}{x}$

$\Rightarrow 9{{x}^{2}}+63x-162=0$

$\Rightarrow 9\left( {{x}^{2}}+9x-2x-18 \right)=0$

$\Rightarrow x\left( x+9 \right)-2\left( x+9 \right)=0$

$\Rightarrow x=2\,or\,x=-9$

Number$=10\left( \dfrac{18}{2} \right)+2=92$

11. A factory kept increasing its output by the same percent ago every year. Find the percentage if it is known that the output is doubled in the last two years.

According to question,

2P$=p{{\left( 1+\dfrac{r}{100} \right)}^{2}}$

$\Rightarrow \dfrac{\sqrt{2}}{1}=1+\dfrac{r}{100}$

$\sqrt{2}-1=\dfrac{r}{100}$

$\Rightarrow r=\left( \sqrt{2}-1 \right)100$

12. Two pipes running together can fill a cistern in if one pipe takes $3\dfrac{1}{13}$ minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Let the faster pipe takes minutes to fill the cistern and the slower pipe will take $\left( x+3 \right)$ minutes.

$\dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{1}{\dfrac{40}{13}}$

$\Rightarrow \dfrac{1}{x}+\dfrac{1}{x+3}=\dfrac{13}{40}$

$\Rightarrow \dfrac{x+3+x}{{{x}^{2}}+3x}=\dfrac{13}{40}$

$\Rightarrow 13{{x}^{2}}-41x-120=0$

$\Rightarrow 13{{x}^{2}}-65x+24x-120=0$

$\Rightarrow 13x\left( x-5 \right)+24\left( x-5 \right)=0$

$\therefore x=5\,or\,x=\dfrac{-24}{13}\left( Neglected \right)$

13. If the roots of the equation$\left( a-b \right){{x}^{2}}+\left( b-c \right)x+\left( c-a \right)=0$ are equal, prove that$2a=b+c$

$\left( a-b \right){{x}^{2}}+\left( b-c \right)x+\left( c-a \right)=0$

$={{\left( b-c \right)}^{2}}-4\times \left( a-b \right)\times \left( c-a \right)$

$={{b}^{2}}+{{c}^{2}}-2bc-4\left( ac-{{a}^{2}}-bc+ab \right)$

$={{b}^{2}}+{{c}^{2}}-2bc-4ac+4{{a}^{2}}+4bc-4ab$

$={{\left( b \right)}^{2}}+{{\left( c \right)}^{2}}+{{\left( 2a \right)}^{2}}+2bc-4ac-4ab$

$={{\left( b+c-2a \right)}^{2}}$

For equal root s,

D = 0

$\Rightarrow {{\left( b+c-2a \right)}^{2}}=0$

$\Rightarrow \left( b+c-2a \right)=0$

$\Rightarrow b+c=2a$

14. Two circles touch internally. The sum of their areas is $116\pi c{{m}^{2}}$ and the distance between their centers is $6$ cm. Find the radii of the circles.

Let ${{r}_{1}}$ and ${{r}_{2}}$ be the radius of two circles

$\Pi {{r}_{1}}^{2}+\Pi {{r}_{2}}^{2}=116\Pi $

$\Rightarrow {{r}_{1}}^{2}+{{r}_{2}}^{2}=116......\left( i \right)$

${{r}_{2}}-{{r}_{1}}=6$ (Given)

$\Rightarrow {{r}_{2}}=6+{{r}_{1}}$

Puting the value of${{r}_{2}}$ in eq. … (i) we get

$\Rightarrow {{r}_{1}}^{2}+36+{{r}_{1}}^{2}+12{{r}_{1}}=116$

$\Rightarrow 2{{r}_{1}}^{2}+12{{r}_{1}}-80=0$

${{r}_{1}}^{2}+6{{r}_{1}}-40=0$

$\Rightarrow {{r}_{1}}^{2}+10{{r}_{1}}-4{{r}_{1}}-40=0$

$\Rightarrow {{r}_{1}}\left( {{r}_{1}}+10 \right)-4\left( {{r}_{1}}+10 \right)=0$

$\Rightarrow \left( {{r}_{1}}+10 \right)\left( {{r}_{1}}-4 \right)=0$

$when\,{{r}_{1}}=4\,cm$

${{r}_{2}}=6+{{r}_{1}}$

$=6+4$

${{r}_{2}}=10\,cm$

$\Rightarrow {{r}_{1}}=-10$ (Neglect) or ${{r}_{1}}=\text{ }4\text{ }cm$

15. A piece of cloth costs Rs.$200$. If the piece was $5$ m longer and each metre of cloth costs Rs. $2$ less the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter?

Let the length of piece $=x$ m

Rate per meter $=\dfrac{200}{x}$

A New length $=\text{ }\left( x+5 \right)$

A New rate per meter $=\dfrac{200}{x+5}$

$\dfrac{200}{x}-\dfrac{200}{x+5}=2$

$\Rightarrow \dfrac{200\left( x+5 \right)-200x}{\left( x+5 \right)}=\dfrac{2}{1}$

$\Rightarrow \dfrac{200x+1000-200x}{{{x}^{2}}+5x}=\dfrac{2}{1}$

$\Rightarrow {{x}^{2}}+5x=500$

$\Rightarrow {{x}^{2}}+25x-20x-500=0$

$\Rightarrow x\left( x+25 \right)-20\left( x+25 \right)=0$

$\Rightarrow \left( x+25 \right)\left( x-20 \right)=0$

$\therefore x=-25\left( Neglect \right)\,or\,x=20$

Rate per meter $=\text{ }10$

16. $a{{x}^{2}}+bx+x\text{ }=\text{ }0$ , $a\ne 0$ solve by quadratic formula.

$a{{x}^{2}}+bx+x\text{ }=\text{ }0$

$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x+\dfrac{c}{a}=0$

$\Rightarrow {{x}^{2}}+\dfrac{b}{a}x{{\left( \dfrac{b}{2a} \right)}^{2}}-{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{c}{a}=0$

$\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}}{4{{a}^{2}}}-\dfrac{c}{a}$

x$\Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$

$\Rightarrow x+\dfrac{b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}}$

$\Rightarrow x+\dfrac{-b}{2a}=\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{2a}}$

$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\Rightarrow x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$

$or\,\Rightarrow x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$

17. The length of the hypotenuse of a right-angled triangle exceeds the length of the base by $2$ cm and exceeds twice the length of the altitude by $1$ cm. Find the length of each side of the triangle.

Let base of the triangle $=\text{ }x$

Altitude of the triangle $=\text{ }y$

Hypotenuse of the triangle $=\text{ }h$

$h=x+2$

$h=2y+1$

$\Rightarrow x+2=2y+1$

$\Rightarrow x+2-1=2y$

$\Rightarrow x-1=2y$

$\Rightarrow \dfrac{x-1}{2}=y$

$And\,{{x}^{2}}+{{y}^{2}}={{h}^{2}}$

$\Rightarrow {{x}^{2}}+{{\left( \dfrac{x-1}{2} \right)}^{2}}={{\left( x+2 \right)}^{2}}$

$\Rightarrow {{x}^{2}}-15x+x-15=0$

$\Rightarrow \left( x-15 \right)\left( x+1 \right)=0$

$\Rightarrow x=15\,or\,x=-1$

Base of the triangle $=\text{ }15$ cm

Altitude of the triangle $=\dfrac{x+1}{2}=8cm$

Hypotenuse of the triangle $=\text{ }17$ cm

18. Find the roots of the following quadratic equations if they exist by the method of completing square.

i). $2{{x}^{2}}\text{ }-7x+3=0$

(i) $2{{x}^{2}}\text{ }-7x+3=0$

Dividing the equation by $2$ to make coefficient of ${{x}^{2}}$ equal to$1$ we get

${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$

Dividing the middle term of the equation by$2x$, we get

$\dfrac{7}{2}x\times \dfrac{1}{2x}=\dfrac{7}{4}$

Adding and subtracting square of $\dfrac{7}{4}$ from the equation ${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}=0$ we get

${{x}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+{{\left( \dfrac{7}{4} \right)}^{2}}-{{\left( \dfrac{7}{4} \right)}^{2}}=0$

$\Rightarrow {{x}^{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}-\dfrac{7}{2}x+\dfrac{3}{2}+-{{\left( \dfrac{7}{4} \right)}^{2}}=0$ $\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right\}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{+}\dfrac{\text{24-49}}{\text{16}}\text{=0}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49-24}}{\text{16}}$

Square rooting on both the sides we get

$\Rightarrow \text{x-}\dfrac{\text{7}}{\text{4}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\text{5}}{\text{4}}$

$\Rightarrow x\text{=}\dfrac{\text{5}}{\text{4}}\text{+}\dfrac{\text{7}}{\text{4}}\text{=}\dfrac{\text{12}}{\text{4}}\text{=3}\,\text{and}\,\text{x=-}\dfrac{\text{5}}{\text{4}}\text{+}\dfrac{\text{7}}{\text{4}}\text{=}\dfrac{\text{2}}{\text{4}}\text{=}\dfrac{\text{1}}{\text{2}}$

Therefore,$x=\dfrac{1}{2},3$

ii). $2{{x}^{2}}+x\text{ }4=0$

$\text{2}{{\text{x}}^{\text{2}}}\text{+x-- 4=0}$

Dividing equation by$2$ ,

${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{-2=0}$

Following procedure of completing square,

${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{-2+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-2-}\dfrac{\text{1}}{\text{16}}\text{=0}$ $\left\{ {{\left( \text{a-b} \right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{-2ab} \right\}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}\dfrac{\text{33}}{\text{16}}\text{=0}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{-}\dfrac{\text{33}}{\text{16}}$

Taking square root on both sides,

$\Rightarrow \text{x+}\dfrac{\text{1}}{\text{4}}\text{= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}$

$\Rightarrow x=\dfrac{\sqrt{33}}{4}-\dfrac{1}{4}\text{=}\dfrac{\sqrt{\text{33}}\text{-1}}{\text{4}}\,\text{and}\,\text{x=-}\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\text{=}\dfrac{\text{-}\sqrt{\text{33}}\text{-1}}{\text{4}}$

Therefore,$\text{x}=\dfrac{\sqrt{33}-1}{4},\dfrac{-\sqrt{33}-1}{4}$

$\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{\text{3}}\text{x+3=0}$

Dividing this equation by $4,$ we get

${{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{=0}$

By the procedure of completing square we get

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{+}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{-}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{+}\sqrt{\text{3}}\text{x+}\dfrac{\text{3}}{\text{4}}\text{-}\dfrac{\text{3}}{\text{4}}\text{=0}$ $\left\{ {{\left( \text{a-b} \right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{-2ab} \right\}$

$\Rightarrow {{\left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)}^{\text{2}}}\text{=0}$

$\Rightarrow \left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)\left( \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}} \right)\text{=0}$

$\Rightarrow \text{x+}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{=0,x+}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{=0}$

$\Rightarrow \text{x=-}\dfrac{\sqrt{\text{3}}}{\text{2}}\text{,-}\dfrac{\sqrt{\text{3}}}{\text{2}}$

iv). $2{{x}^{2}}+x+4=0$

$\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Dividing this equation by$2$we get

${{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+2=0}$

By the procedure of completing square,

$\Rightarrow {{x}^{2}}+\dfrac{x}{2}+2+{{\left( \dfrac{1}{4} \right)}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{+}\dfrac{\text{x}}{\text{2}}\text{+2-}{{\left( \dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=0}$ $\left\{ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \right\}$

$\Rightarrow {{\left( x+\dfrac{1}{4} \right)}^{2}}+2-\dfrac{1}{16}=0$

${{\left( x+\dfrac{1}{4} \right)}^{2}}=\dfrac{1}{16}-2=\dfrac{1-32}{16}$

Right hand side does not exist because the square root of the negative number does not exist.

Therefore, there is no solution for quadratic equation $2{{x}^{2}}+x+4=0$

Practice Questions for Class 10 Maths Chapter 4 - Quadratic Equations

The following are some of the questions that can be taken up by students to assist them in the board preparations related to Quadratic Equations.

Question 1. A rectangular field's diagonal is 60 metres longer than the shorter side. Find the field's sides if the longer side is 30 metres more than the shorter side.

Anwer- 120m which is the correct answer.

Question 2. Is it possible to build a rectangular park with an 80 meter perimeter and a 400-square-meter area? If this is the case, determine its length and breadth.

Answer- Length and breadth both should be equal to 20 m.

Question 3- A train moving at a speed of 600 km was slowed down due to bad weather. The train's average speed dropped by 200 km/hr, and the journey time was increased by 30 minutes. Determine the train's initial duration.

Answer- 1 hour is the answer.

Vedantu's goal is to help students from all over the country prepare for exams. As a result, all of our study materials are available in PDF format, which can be downloaded for free. Our professionals answer the questions to provide students with a sample answer for each question on the Class 10 CBSE test papers . For practice, students can download and solve the question paper on their own. For exam preparation, they can use Vedantu to access critical questions, revision notes, NCERT chapter-by-chapter solutions , and other resources.

Important Related Links for CBSE Class 10 Maths

FAQs on Important Questions for CBSE Class 10 Maths Chapter 4 - Quadratic Equations

Q1: What are the roots of a quadratic equation? Explain the nature of roots of a quadratic equation.

A1: The roots of the quadratic equation are given as : x = (-b ± √D)/2a, where D = b 2 – 4ac.

The nature of the roots of a quadratic equation are as mentioned below:

D > 0, roots are real and distinct (unequal)

D = 0, roots are real and equal (coincident)

D < 0, roots are imaginary and unequal

Q2: Explain in detail the nature of roots of a quadratic equation.

A2: The nature of roots of a quadratic equation is as follows:

If the value of discriminant = 0 i.e. b 2 – 4ac = 0, then the quadratic equation will have equal roots i.e. α = β = -b/2a.

If the value of discriminant < 0 i.e. b 2 – 4ac < 0, then the quadratic equation will have imaginary roots i.e α = (p + iq) and β = (p – iq). Where ‘iq’ is the imaginary part of a complex number.

If the value of discriminant (D) > 0 i.e. b 2 – 4ac > 0, then the quadratic equation will have real roots.

If the value of discriminant > 0 and D is a perfect square, then the quadratic equation will have rational roots.

If the value of discriminant (D) > 0 and D is not a perfect square, then the quadratic equation will have irrational roots i.e. α = (p + √q) and β=(p – √q).

If the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers, then the quadratic equation will have integral roots.

Q3: What is the Relationship between Coefficient and Roots of Quadratic Equation?

A3: If α and β are roots of a Quadratic Equation ax 2 + bx + c then,

α + β = -b/a

α – β = ±√[(α + β) 2 – 4αβ]

|α + β| = √D/|a|

The relationship between the roots and coefficient of a polynomial equation can be derived by simplifying the given polynomials and substituting the above results as shown below.

α 2 β + β 2 α = αβ (α + β) = – bc/a 2

α 2 + αβ + β 2 = (α + β) 2 – αβ = (b 2 – ac)/a 2

α 2 + β 2 = (α – β) 2 – 2αβ

α 2 – β 2 = (α + β) (α – β)

α 3 + β 3 = (α + β) 3 + 3αβ(α + β)

α 3 – β 3 = (α – β) 3 + 3αβ(α – β)

(α/β) 2 + (β/α) 2 = α 4 + β 4 /α 2 β 2

Q4: What happens when a < 0 and b 2 – 4ac > 0.

A4: When a < 0 and b 2 – 4ac > 0, the graph of a quadratic equation will be concave downwards and will intersect the x-axis at two points α and β with α < β. The quadratic equation will have two real roots (α and β) and the curve will always lie below the x-axis.

The quadratic function f(x) will be positive i.e. f(x) > 0, for the values of x lying in the interval (α, β).

The quadratic function f(x) will be equal to zero i.e. f(x) = 0, if x = α or β

The quadratic function f(x) will be negative i.e. f(x) < 0 for the values of x lying in the interval (−∞, α) ∪ (β, ∞).

CBSE Class 10 Maths Important Questions

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CBSE Class 10 Maths Case Study

CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus.

These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.

Class 10 Maths (Formula, Case Based, MCQ, Assertion Reason Question with Solutions)

In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.

Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too.

Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.

For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.

CBSE Class 10th Case study Questions

Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.

Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.

CBSE Class 10th Assertion Reasoning Questions

These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement?

Moreover, to solve the problem they need to look at the given options and then answer them.

CBSE Class 10 Maths Case Based MCQ

CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.

Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.

Class 10th Mathematics Multiple Choice Questions

Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.

Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.

The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.

To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.

In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.

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Important Questions for Class 10 Maths Chapter 4 Quadratic Equations

Chapter 4 quadratic equations important questions for cbse class 10 maths board exams.

Important Questions for Chapter 4 Quadratic Equations Class 10 Maths

Quadratic equations class 10 maths important questions very short answer (1 mark).

Let one root be α and other root be 6α.

Since k = 0 is not possible, therefore k = 3.

4. If 1 is a root of the equations ay 2 + ay + 3 = 0 and y 2 + y + b = 0, then find the value of ab.

ay 2 + ay + 3 = 0 ⇒ a(1) 2 + a(1) + 3 = 0 ⇒ 2a = -3 ⇒ a = −32

y 2 + y + b = 0 ⇒ 1 2 + 1 + b = 0 ⇒ b = -2 ∴ ab = (−32)(−2) = 3

5. Find the roots of the equation x 2 – 3x – m (m + 3) = 0, where m is a constant.

x 2 – 3x – m(m + 3) = 0 ⇒ D = b 2 – 4ac ⇒ D = (- 3) 2 – 4(1) [-m(m + 3)] = 9 + 4m (m + 3) = 4m 2 + 12m + 9 = (2m + 3) 2

∴ x = m + 3 or -m

Quadratic Equations Class 10 Maths Important Questions Short Answer (2 Marks)

6. For what values of k, the roots of the equation x 2 + 4x + k = 0 are real?

Comparing the given equation with ax 2 + bx + c = 0, we get a = 1, b = 4, c = k.

Since, given the equation has real roots,

D ≥ 0

⇒ b 2 - 4ac ≥ 0

⇒ 4 2 - 4 × 1 × k ≥ 0

⇒ 4k ≤ 16

⇒ k ≤ 4

7. Find the value of m so that the quadratic equation mx (x – 7) + 49 = 0 has two equal roots.

We have, mx (x – 7) + 49 = 0 mx 2 – 7mx + 49 = 0 Here, a = m, b = – 7m, c = 49 D = b 2 – 4ac = 0 …[For equal roots] ⇒ (-7m) 2 – 4(m) (49) = 0 ⇒ 49m 2 – 4m (49) = 0 ⇒ 49m (m – 4) = 0 ⇒ 49m = 0 or m – 4 = 0 m = 0 (rejected) or m = 4 ∴ m = 4

8. Find the value of k for which the roots of the equations 3x 2 - 10x + k = 0 are reciprocal of each other.

Comparing the given equation with ax 2 + bx + c = 0 we get a = 3, b = -10, c = k

Let one root be α so other root is 1/α

product of roots, α × 1/α = c/a

Hence, the value of k is 3.

9. Find the roots of the following quadratic equation :

15x 2 - 10√6x + 10 = 0

We have 15x 2 - 10√6x + 10 = 0

We have, px (x – 3) + 9 = 0 px 2 – 3px + 9 = 0 Here a = p, b = -3p, D = 0 b 2 – 4ac = 0 ⇒ (-3p) 2 – 4(p)(9) = 0 ⇒ 9p 2 – 36p = 0 ⇒ 9p (p – 4) = 0 ⇒ 9p = 0 or p – 4= 0 p = 0 (rejected) or p = 4 ∴ p = 4 …(∵ Coefficient of x 2 cannot be zero)

11. Solve for x:

36x 2 – 12ax + (a 2 – b 2 ) = 0

We have, 36x 2 – 12ax + (a 2 – b 2 ) = 0 ⇒ (36x 2 – 12ax + a 2 ) – b 2 = 0 ⇒ [(6x) 2 – 2(6x)(a) + (a) 2 ] – b 2 = 0 ⇒ (6x – a) 2 – (b) 2 = 0 …[∵ x 2 – 2xy + y 2 = (x – y) 2 ] ⇒ (6x – a + b) (6x – a – b) = 0 ...[∵ x 2 – y 2 = (x + y)(x – y)] ⇒ 6x – a + b = 0 or 6x – a – b = 0 ⇒ 6x = a – b or 6x = a + b ⇒ x = a− b/6 or a+ b/6

12. Solve for x:

4√3x 2 + 5x - 2√3 = 0

Here 4x 2 + 3x + 5 = 0

But (2x+34) 2 cannot be negative for any real value of x.

14. Solve for x : x 2 - (√3+1)x + √3 = 0

x 2 - (√3+1)x + √3 = 0

⇒ x 2 -√3 x - 1x + √3 = 0

⇒ x(x-√3) - 1(x-√3) = 0

⇒ (x-√3) (x-1) = 0

Thus, x = √3, x = 1

15. Find the roots of the following quadratic equation :

(x+3)(x-1) = 3(x - 1/3)

9x 2 - 6b 2 x - (a 4 - b 4 ) = 0

Comparing with Ax 2 + Bx + C

A = 9, B = -6b 2 , C = -(a 4 - b 4 )

We have, ax 2 + 7x + b = 0 Here ‘a’ = a, ‘b’ = 7, ‘c’ = b Now, α = 23 and β = -3 …[Given]

The given quadratic equation can be written as (9x 2 – 6b 2 x + b 4 ) – a 4 = 0 ⇒ (3x – b 2 ) 2 – (a 2 ) 2 = 0 ⇒ (3x – b 2 + a 2 ) (3x – b 2 – a 2 ) = 0 …[x 2 – y 2 = (x + y) (x – y)] ⇒ 3x – b 2 + a 2 = 0 or 3x – b 2 – a 2 = 0 ⇒ 3x = b 2 – a 2 or 3x = b 2 + a 2

We have, 2x 2 + px – 15 =0 Since (-5) is a root of the given equation ∴ 2(-5) 2 + p(-5) – 15 = 0 ⇒ 2(25) – 5p – 15 = 0 ⇒ 50 – 15 = 5p ⇒ 35 = 5p ⇒ p = 7 …(i)

Now, p(x 2 + x) + k ⇒ px 2 + px + k = 0 7x 2 + 7x + k = 0 …[From (i)] Here, a = 7, b = 7, c = k

D = 0 …[Roots are equal] ⇒ b 2 – 4ac = 0 ⇒ (7) 2 – 4(7)k = 0 ⇒ 49 – 28k = 0 ⇒ 49 = 28k ∴ k = 49/28 = 7/4

13√3 x 2 + 10x + √3 = 0

Comparing with ax2 + bx + c = 0, we get

a = 13√3, b = 10, c = √3

b 2 - 4ac = 10 2 - 4(13√3)(√3)

As D < 0, the equation has no real roots.

22. Find the positive value of k for which x 2 - 8x + k = 0, will have real roots.

x 2 - 8x + k = 0

Comparing with Ax 2 + Bx + C = 0, we get

A = 1, B = -8, C= k

Since the given equation has real roots,

B 2 - 4AC > 0

⇒ (-8) 2 - 4(1)(k) ≥ 0

⇒ 64 - 4k ≥ 0

⇒ 16 - k ≥ 0

⇒ 16 ≥ k

Thus, k ≤ 16

23. If 2 is a root of the equation x 2 + kx + 12 = 0 and the equation x 2 + kx + q = 0 has equal roots, find the value of q.

x 2 + kx + 12 = 0

If 2 is the root of above equation, it must satisfy it.

22 + 2k + 12 = 0

⇒ 2k + 16 = 0

⇒ k = -8

Substituting k = -8 in x 2 + kx + q = 0 we have

x 2 - 8x + q = 0

For equal roots,

(-8) 2 - 4(1)q = 0

⇒ 64 - 4q = 0

⇒ 4q = 64

⇒ q = 16

24. Solve for x : √3 x2 + 10x + 7√3 = 0

2x 2 + kx + 8 = 0

Comparing with ax 2 + bx + c = 0, we get

a = 2, b = k and c = 8

For equal roots, D = 0,

b 2 - 4ac = 0

k 2 - 4×2×8 = 0

⇒ k 2 = 64

⇒ k = ±√64

Thus k = ±8

Quadratic Equations Class 10 Maths Important Questions Short Answer-II (3 Marks)

26. Find the values of k for which the quadratic equation x 2 + 2√2 kx + 18 = 0 has equal roots.

x 2 + 2√2 kx + 18 = 0

Comparing it by ax 2 + bx + c, we get a = 1, b = 2√2 k and c = 18.

Given that,

Equation x 2 + 2√2 kx + 18 = 0 has equal roots.

⇒ (2√2 k) 2 -4118 = 0

⇒ 8k 2 - 72 = 0

⇒ 8k 2 = 72

⇒ k 2 = 72/8 = 9

⇒ k = ±3

27. If α and β are the zeroes of the polynomial f(x) = x 2 - 4x - 5 then find the value of α 2 + β 2 .

p(x) = x 2 - 4x - 5

Comparing it by ax 2 + bx + c, we get a = 1, b = -4 and c = -5.

Since, given α and β are the zeroes of the polynomial,

28. Find the quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Hence find the zeroes.

Sum of zeroes,

α + β = -3 ...(1)

and product of zeroes, αβ = 2

Thus quadratic equation is

x 2 - (α+β)x + αβ = 0

⇒ x 2 - (-3)x + 2 = 0

⇒ x 2 + 3x + 2 = 0

Thus quadratic equation is x 2 + 3x + 2 = 0

Now, above equation can be written as

x 2 + 2x + x + 2 = 0

⇒ x(x+2) + (x+2) = 0

⇒ (x+2) (x+1) = 0

Hence, zeroes are -2 and -1.

29. Find the roots of the following quadratic equation: 2√3 x 2 – 5x + √3 = 0

We have, 2√3 x 2 – 5x + √3 = 0

Here, a = 2√3, b = -5, c = √3

D = b 2 – 4ac ∴ D = (-5) 2 – 4 (2√3)(√3) = 25 – 24 = 1

31. Solve for x: 4x 2 – 4ax + (a 2 – b 2 ) = 0

33. Find the value(s) of k so that the quadratic equation 3x 2 – 2kx + 12 = 0 has equal roots.

Given: 3x 2 – 2kx + 12 = 0 Here a = 3, b = -2k, c = 12 D = 0 …[Since roots are equal as b 2 – 4ac = 0] ∴ (-2k) 2 – 4(3) (12) = 0 ⇒ 4k 2 – 144 = 0 ⇒ k 2 = 144/4 = 36 ∴ k = ±√36 ∴ k = ±6

34. Find the quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Hence find the zeroes.

Sum of zeroes α + β = -3

and product of zeroes αβ = 2

Thus, quadratic equation is x 2 + 3x + 2 = 0

35. Find the zeroes of the quadratic polynomial 6x 2 - 3 - 7x and verify the relationship between the zeroes and the coefficients.

We have, p(x) = 6x 2 - 3 - 7x For zeroes of polynomial, p(x) = 0 6x 2 - 7x - 3 = 0 ⇒ 6x 2 - 9x + 2x - 3 = 0 ⇒ 3x(2x - 3) + 1(2x-3) = 0 ⇒ (2x-3) (3x+1) = 0 Thus, 2x - 3 = 0 and 3x + 1 = 0 Hence, x = 3/2 and x = -1/3 Therefore, α = 3/2 and β = -1/3 are the zeroes of the given polynomial. Verification:

We have, (k + 4) x 2 + (k + 1) x + 1 = 0 Here, a = k + 4, b = k + 1, c = 1 D =0 …[∵ Roots are equal] b 2 – 4ac = 0 ∴ (k + 1) 2 – 4(k + 4)(1) = 0 k 2 + 2k + 1 – 4k – 16 = 0 ⇒ k 2 – 2k – 15 = 0 ⇒ k 2 – 5k + 3k – 15 = 0 ⇒ k(k – 5) + 3(k – 5) = 0 ⇒ (k – 5)(k + 3) = 0 ⇒ k – 5 = 0 or k + 3= 0 ⇒ k = 5 or k = -3 ∴ k = 5 and -3

38. Find the zeroes of the quadratic polynomial x 2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.

p(x) = x 2 + 7x + 10

For zeroes of polynomial p(x) = 0,

x 2 + 7x + 10 = 0

⇒ x 2 + 5x + 2x + 10 = 0

⇒ x(x+5) + 2(x+5) = 0

⇒ (x+5) (X+2) = 0

x = -2 and x = -5

Therefore, α = -2 and β = -5 are the zeroes of the given polynomial.

Verification:

α + β = -2 + (-5)

y 2 + k 2 = 2(k + 1)y y 2 – 2(k + 1)y + k 2 = 0 Here a = 1, b = -2(k + 1), c = k 2 D = 0 …[Roots are equal] b 2 – 4ac = 0 ∴ [-2(k + 1)] 2 – 4 × (1) × (k 2 ) = 0 ⇒ 4(k 2 + 2k + 1) – 4k 2 = 0 ⇒ 4k 2 + 8k + 4 – 4k 2 = 0 ⇒ 8k + 4 = 0 ⇒ 8k = -4 ∴ k = -4/8 = -1/2

41. Solve the equation 3/x+1 - 1/2 = 2/3x-1, x≠ -1, x≠ 1/3 for x.

⇒ 2(2x + 2) = (5 – x)(3x – 1) ⇒ 4x + 4 = 15x – 5 – 3x 2 + x ⇒ 4x + 4 – 15x + 5 + 3x 2 – x = 0 ⇒ 3x 2 – 12x + 9 = 0 ⇒ x 2 – 4x + 3 = 0 …[Dividing by 3] ⇒ x 2 – 3x – x + 3 = 0 ⇒ x(x – 3) – 1(x – 3) = 0 ⇒ (x – 1) (x – 3) = 0 ⇒ x – 1 = 0 or x – 3 = 0 ∴ x = 1 or x = 3

42. Solve for x:

2x/x-3 + 1/2x+3 + 3x+9/(x+9)(2x+3) = 0, x ≠ 3, -3/2

2x(2x+3) + (x-3) + (3x+9) = 0 ⇒ 4x 2 + 6x + x - 3 + 3x + 9 = 0 ⇒ 4x 2 + 10x + 6 = 0 ⇒ 2x 2 + 5x + 3 =0 ⇒ (x+1) (2x+3) = 0 Thus, x = -1, x = -3/2

43. Solve the following quadratic equation for x :

x 2 2 + (a/a+b + a+b/a)x + 1 = 0

x 2 - (2b-1)x + (b 2 -b-20) = 0

Comparing with Ax 2 + Bx + C = 0 we have

A = 1, B = -(2b-1), C = (b 2 -b-20)

Thus x = b+4 and x = b-5

46. Find the roots of the equation 2x 2 + x - 4 = 0, by the method of completing the squares.

2x 2 + x - 4 = 0

Let three consecutive natural numbers are x, x + 1, x + 2. According to the question, (x + 1) 2 – [(x + 2) 2 – x 2 ] = 60 ⇒ x 2 + 2x + 1 – (x 2 + 4x + 4 – x 2 ) = 60 ⇒ x 2 + 2x + 1 – 4x – 4 – 60 = 0 ⇒ x 2 – 2x – 63 = 0 ⇒ x 2 – 9x + 7x – 63 = 0 ⇒ x(x – 9) + 7(x – 9) = 0 ⇒ (x – 9) (x + 7) = 0 ⇒ x – 9 = 0 or x + 7 = 0 ⇒ x = 9 or x = -7 Natural nos. can not be negative, ∴ x = 9 ∴ Numbers are 9, 10, 11.

48. If the sum of two natural numbers is 8 and their product is 15, find the numbers.

Let the numbers be x and (8 – x). According to the Question, x(8 – x) = 15 ⇒ 8x – x 2 = 15 ⇒ 0 = x 2 – 8x + 15 ⇒ x 2 – 5x – 3x + 15 = 0 ⇒ x(x – 5) – 3(x – 5) = 0 ⇒ (x – 3)(x – 5) = 0 x – 3 = 0 or x – 5 = 0 x = 3 or x = 5 When x = 3, numbers are 3 and 5. When x = 5, numbers are 5 and 3.

49. If 2 is a root of the quadratic equation 3x 2 + px - 8 = 0 and the quadratic equation 4x 2 - 2px + k = 0 has equal roots, find k.

3x 2 + px - 8 = 0

Since 2 is a root of above equation, it must satisfy it.

Substituting, x= 2 in we have

12 + 2p - 8 = 0

⇒ p = -2

4x 2 - 2px + k = 0 has equal roots.

or, 4x 2 + 4x + k = 0 has equal roots.

D = b 2 - 4ac = 0

4 2 - 4(4)(k) = 0

⇒ 16 - 16k = 0

⇒ 16k = 16

⇒ k = 1

50. If -3 is a root of quadratic equation 2x 2 + px - 15 = 0, while the quadratic equation x 2 - 4px + k = 0 has equal roots. Find the value of k.

Given -3 is a root of quadratic equation.

2x 2 + px - 15

Since 3 is a root of above equation, it must satisfy it.

Substituting x 3 = in above equation we have,

2(-3) 2 + p(-3) - 15 = 0

⇒ 2×9 - 3p - 15 = 0

⇒ p = 1

x 2 - 4px + k = 0 has equal roots

or, x 2 - 4x + k = 0 has equal roots.

⇒ (-4) 2 - k = 0

⇒ 16 - 4k = 0

⇒ 4k = 16

⇒ k = 4

51. Solve 1/(a+b+x) = 1/a + 1/b + 1/x, a+b ≠ 0

Quadratic Equations Class 10 Maths Important Questions Long Answer (4 Marks)

⇒ 2x = -8x + 40

⇒ 10x = 40

⇒ x = 4

Hence, x = 15, 4

53. Solve for x: 1/x+1 + 2/x+2 = 4/x+4, x≠ -1,-2,-4

(3k + 1)x 2 + 2(k + 1) + 1 = 0 Here, a = 3k + 1, b = 2(k + 1), c = 1 D = 0 …[∵ Roots are equal] As b 2 – 4ac = 0 ∴ [2(k + 1)] 2 – 4(3k + 1)(1) = 0 ⇒ 4(k + 1) 2 – 4(3k + 1) = 0 ⇒ 4(k 2 + 2k + 1 – 3k – 1) = 0 ⇒ (k 2 – k) = 0/4 ⇒ k(k – 1) = 0 ⇒ k = 0 or k – 1 = 0 ∴ k = 0 or k = 1 Roots are x = −b/2a ...[ Given, equal roots]

(2p + 1)x 2 – (7p + 2)x + (7p – 3) = 0 Here, a = 2p + 1, b = -(7p + 2), c = 7p – 3 D = 0 …[∵ Equal roots As h2 – 4ac = 0] ∴ [-(7p + 2)] 2 – 4(2p + 1)(7p – 3) = 0 ⇒ (7p + 2) 2 – 4(14p 2 – 6p + 7p – 3) = 0 ⇒ 49p 2 + 28p + 4 – 56p 2 + 24p – 28p + 12 = 0 ⇒ -7p 2 + 24p + 16 = 0 ⇒ 7p 2 – 24p – 16 = 0 …[Dividing both sides by -1] ⇒ 7p 2 – 28p + 4p – 16 = 0 ⇒ 7p(p – 4) + 4(p – 4) = 0 ⇒ (p – 4) (7p + 4) = 0 ⇒ p – 4 = 0 or 7p + 4 = 0 ⇒ p = 4 or p = −4/7

59. Check whether the equation 5x 2 - 6x - 2 = 0 has real roots if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.

5x 2 - 6x - 2 = 0

Comparing with ax 2 + bx + c = 0 we get,

a = 5, b = -6 and c = -2

b 2 - 4ac = (-6) 2 - 4×5×(-2)

= 36 + 40 = 76 > 0

So the equation has real and two distinct roots.

5x 2 - 6x = 2

Dividing both the sides by 5 we get

Let the number of books he bought = x Increased number of books he had bought = x +4 Total amount = ₹80 According to the problem,

⇒ x(x + 4) = 320 ⇒ x 2 + 4x – 320 = 0 ⇒ x 2 + 20x – 16x – 320 = 0 ⇒ x(x + 20) – 16(x + 20) = 0 ⇒ (x + 20) (x – 16) = 0 ⇒ x + 20 = 0 or x – 16 = 0 ⇒ x = -20 …(neglected) or x = 16 ∴ Number of books he bought = 16

62. Find the positive values of k for which quadratic equations x 2 + kx + 64 = 0 and x 2 - 8x + k = 0 both will have the real roots.

(i) For x 2 + kx + 64 = 0 to have real roots

k 2 - 256 ≥ 0

⇒ k 2 ≥ 256

⇒ k ≥ 16 or k < -16

(ii) For x 2 - 8x + k = 0 to have real roots

64 - 4k ≥ 0

Therefore, For (i) and (ii) to hold simultaneously

63. Sum of the areas of two squares is 400 cm 2 . If the difference of their perimeters is 16 cm, find the sides of the two squares.

Let the side of Large square = x cm Let the side of small square = y cm According to the Question, x 2 + y 2 = 400 …(i) …[∵ area of square = (side) 2 ] 4x – 4y = 16 …[∵ Perimeter of square = 4 sides] ⇒ x – y = 4 …[Dividing both sides by 4] ⇒ x = 4 + y …(ii) Putting the value of x in equation (i), (4 + y) 2 + y2 2 = 400 ⇒ y 2 + 8y + 16 + y 2 – 400 = 0 ⇒ 2y 2 + 8y – 384 = 0 ⇒ y 2 + 4y – 192 = 0 …[Dividing both sides by 2] ⇒ y 2 + 16y – 12y – 192 = 0 ⇒ y(y + 16) – 12(y + 16) = 0 ⇒ (y – 12)(y + 16) = 0 ⇒ y – 12 = 0 or y + 16 = 0 ⇒ y = 12 or y = -16 …[Neglecting negative value] ∴ Side of small square = y = 12 cm and Side of large square = x = 4 + 12 = 16 cm

64. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Let the length of shorter side be x m. ∴ length of diagonal = (x + 16) m and length of longer side = (x + 14) m Using pythagoras theorem, (l) 2 + (b) 2 = (h) 2 ∴ x 2 + (x + 14)2 2 = (x + 16) 2 ⇒ x 2 + x 2 + 196 + 28x = x 2 + 256 + 32x ⇒ x 2 – 4x – 60 = 0 ⇒ x 2 – 10x + 6x – 60 = 0 ⇒ x(x – 10) + 6(x – 10) = 0 ⇒ (x – 10) (x + 6) = 0 ⇒ x – 10 = 0 or x + 6 = 0 ⇒ x = 10 or x = -6 (Neglect as length cannot be negative]) ⇒ x = 10 m

Length of shorter side = x = 10 m Length of diagonal = (x + 16) m = 26 m Length of longer side = (x + 14)m = 24m ∴ Length of sides are 10 m and 24 m.

65. The sum of two numbers is 9 and the sum of their reciprocals is 1/2. Find the numbers.

Let the numbers be x and 9 – x. According to the Question,

When x = 3, numbers are 3 and 6. When x = 6, numbers. are 6 and 3.

66. Find the nature of the roots of the quadratic equation 4x 2 + 4√3x + 3 = 0.

4x 2 + 4√3x + 3 = 0

Comparing the given equation with ax 2 + bx + c = 0, we get

a = 4, b = 4√3 and c = 3.

Since, b 2 - 4ac = 0, then roots of the given equation are real and equal.

67. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.

Let three numbers in A.P. are a – d, a, a + d. a – d + a + a + d = 12 ⇒ 3a = 12 ⇒ a = 4 (a – d) 3 + (a) 3 + (a + d) 3 = 288 ⇒ a 3 – 3a 2 d + 3ad 2 – d 3 + a 3 + a 3 + 3a 2 d + 3ad 2 + d 3 = 288 ⇒ 3a 3 + 6ad 2 = 288 ⇒ 3a(a 2 + 2d 2 ) = 288 ⇒ 3 × 4(4 2 + 2d 2 ) = 288 ⇒ (16 + 2d 2 ) = 288/12 ⇒ 2d 2 = 24 – 16 = 8 ⇒ d 2 = 4 ⇒ d = ± 2 When, a = 4, d = 2, numbers are: a – d, a, a + d, i.e., 2, 4, 6 When, a = 4, d = -2, numbers are: a – d, a, a + d, i.e., 6, 4, 2

68. Find the values of k for which the quadratic equations (k+4)x 2 + (k+1)x + 1 = 0 has equal roots. Also, find the roots.

(k+4)x 2 + (k+1)x + 1 = 0

Comparing with Ax 2 + Bx + C = 0, we get

A = (k+4), B = (k+1), C = 1

If roots are equal, B 2 - 4AC = 0

(k+1) 2 - 4(k+4)(1) = 0

⇒ k 2 + 1 + 2k - 4k - 16 = 0

⇒ k 2 - 2k -15 = 0

⇒ (k-5) (k+3) = 0

⇒ k = 5, -3

For k = 5, equation becomes

9x 2 + 6x + 1 = 0

⇒ (3x+1) 2 = 0

⇒ x = -1/3

For k = -3, equation becomes

x 2 - 2x + 1 = 0

⇒ (x-1) 2 = 0

⇒ x = 1

Hence, roots are 1 and -1/3

69. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.

Perimeter of right ∆ = 60 cm …[Given] a + b + c = 60 ⇒ a + b + 25 = 60 ⇒ a + b = 60 – 25 = 35 …(i) In rt. ∆ACB, AC 2 + BC 2 = AB 2 b 2 + a 2 = (25) 2 …[Pythagoras’ theorem] ⇒ a 2 + b 2 = 625 …(ii) From (i), a + b = 35 (a + b) 2 = (35) …[Squaring both sides] ⇒ a 2 + b 2 + 2ab = 1225 ⇒ 625 + 2ab = 1225 …[From (ii)] ⇒ 2ab = 1225 – 625 = 600 ⇒ ab = 300 …(iii) Area of ∆ = 1/2 × base × corresponding altitude = 1/2 × b × a = 1/2 (300) ...[From (iii)] = 150 cm 2

70. The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 1/15. Find the fraction.

Let the denominator be x and the numerator be x – 3. ∴ Fraction =x−3/x New denominator = x + 1 According to the Question,

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Class 10 Maths MCQs
Chapter 4 Quadratic Equations

Class 10 Maths Chapter 4 Quadratic Equations MCQs

Class 10 Maths MCQs for Chapter 4 (Quadratic Equations) are available online here with answers. All these objective questions are prepared as per the latest CBSE syllabus (2022 – 2023) and NCERT guidelines. MCQs for Class 10 Maths Chapter 4 are prepared according to the new exam pattern. Solving these multiple-choice questions will help students to score good marks in the board exams, which they can verify with the help of detailed explanations given here. To get chapter-wise MCQs, click here . Also, find the PDF of MCQs to download here for free.

Class 10 Maths MCQs for Quadratic Equations

CBSE board has released the datasheet for the Class 10 Maths exam. It is advised for students to start revising the chapters, for the exam. Here, we have given multiple-choice questions for Chapter 4 quadratic equations, to help students to solve different types of questions, which could appear in the board exam. They can build their problem-solving capacity and boost their confidence level by practising the questions here. Get important questions for class 10 Maths here at BYJU’S.

Click here to download the PDF of additional MCQs for Practice on Quadratic equations, Chapter of Class 10 Maths along with answer key:

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Students can also get access to Quadratic equations Class 10 Notes here.

Below are the MCQs for Quadratic Equations

1. Equation of (x+1) 2 -x 2 =0 has number of real roots equal to:

Answer: (a) 1

Explanation: (x+1) 2 -x 2 =0

X 2 +2x+1-x 2 = 0

Hence, there is one real root.

2. The roots of 100x 2 – 20x + 1 = 0 is:

(a) 1/20 and 1/20

(b) 1/10 and 1/20

(d) None of the above

Answer: (c) 1/10 and 1/10

Explanation: Given, 100x 2 – 20x + 1=0

100x 2 – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1) 2 = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

3. The sum of two numbers is 27 and product is 182. The numbers are:

(a) 12 and 13

(b) 13 and 14

(d) 13 and 24

Answer: (b) 13 and 14

Explanation: Let x is one number

Another number = 27 – x

Product of two numbers = 182

x(27 – x) = 182

⇒ x 2 – 27x – 182 = 0

⇒ x 2 – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

⇒ x = 13 or x = 14

4. If ½ is a root of the quadratic equation x 2 -mx-5/4=0, then value of m is:

Answer: (b) -2

Explanation: Given x=½ as root of equation x 2 -mx-5/4=0.

(½) 2 – m(½) – 5/4 = 0

¼-m/2-5/4=0

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:

(a) Base=10cm and Altitude=5cm

(b) Base=12cm and Altitude=5cm

(d) Base=12cm and Altitude=10cm

Answer: (b) Base=12cm and Altitude=5cm

Explanation: Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base 2 + Altitude 2 = Hypotenuse 2 (From Pythagoras theorem)

∴ x 2 + (x – 7) 2 = 13 2

By solving the above equation, we get;

⇒ x = 12 or x = – 5

Since the side of the triangle cannot be negative.

Therefore, base = 12cm and altitude = 12-7 = 5cm

6. The roots of quadratic equation 2x 2 + x + 4 = 0 are:

(a) Positive and negative

(b) Both Positive

(d) No real roots

Answer: (d) No real roots

Explanation: 2x 2 + x + 4 = 0

⇒ 2x 2 + x = -4

Dividing the equation by 2, we get

⇒ x 2 + 1/2x = -2

⇒ x 2 + 2 × x × 1/4 = -2

By adding (1/4) 2 to both sides of the equation, we get

⇒ (x) 2 + 2 × x × 1/4 + (1/4) 2 = (1/4) 2 – 2

⇒ (x + 1/4) 2 = 1/16 – 2

⇒ (x + 1/4)2 = -31/16

The square root of negative number is imaginary, therefore, there is no real root for the given equation.

Answer: (b) 3

Explanation:

Hence, we can write, √(6+x) = x

x 2 -3x+2x-6=0

x(x-3)+2(x-3)=0

(x+2) (x-3) = 0

Since, x cannot be negative, therefore, x=3

8. The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3. The present age of Rehman is:

Answer: (a) 7

Explanation: Let, x is the present age of Rehman

Three years ago his age = x – 3

Five years later his age = x + 5

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/x-3 + 1/x-5 = 1/3

(x+5+x-3)/(x-3)(x+5) = 1/3

(2x+2)/(x-3)(x+5) = 1/3

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x 2 + 2x – 15

⇒ x 2 – 4x – 21 = 0

⇒ x 2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

We know age cannot be negative, hence the answer is 7.

9. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

(a) 30 km/hr

(b) 40 km/hr

(d) 60 km/hr

Answer: (b) 40 km/hr

Explanation: Let x km/hr be the speed of train.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x 2 + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

Negative value is not considered for speed hence the answer is 40km/hr.

10. If one root of equation 4x 2 -2x+k-4=0 is reciprocal of the other. The value of k is:

Answer: (b) 8

Explanation: If one root is reciprocal of others, then the product of roots will be:

α x 1/α = (k-4)/4

11. Which one of the following is not a quadratic equation?

(a) (x + 2) 2 = 2(x + 3)

(b) x 2 + 3x = (–1) (1 – 3x) 2

(d) x 3 – x 2 + 2x + 1 = (x + 1) 3

Answer: (c) (x + 2) (x – 1) = x 2 – 2x – 3

We know that the degree of a quadratic equation is 2.

By verifying the options,

x 2 + 4x + 4 = 2x + 6

x 2 + 2x – 2 = 0

This is a quadratic equation.

x 2 + 3x = -1(1 + 9x 2 – 6x)

x 2 + 3x + 1 + 9x 2 – 6x = 0

10x 2 – 3x + 1 = 0

x 2 + x – 2 = x 2 – 2x – 3

x 2 + x – 2 – x 2 + 2x + 3 = 0

This is not a quadratic equation.

12. Which of the following equations has 2 as a root?

(a) x 2 – 4x + 5 = 0

(b) x 2 + 3x – 12 = 0

(d) 3x 2 – 6x – 2 = 0

Answer: (c) 2x 2 – 7x + 6 = 0

If 2 is a root then substituting the value 2 in place of x should satisfy the equation.

Let us verify the given options.

(a) x 2 – 4x + 5 = 0

(2) 2 – 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x 2 – 4x + 5 = 0

(2) 2 + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x 2 + 3x – 12 = 0

2(2) 2 – 7(2) + 6 = 0

Here, x = 2 is a root of 2x 2 – 7x + 6 = 0

13. A quadratic equation ax 2 + bx + c = 0 has no real roots, if

(a) b 2 – 4ac > 0

(b) b 2 – 4ac = 0

(d) b 2 – ac < 0

Answer: (c) b 2 – 4ac < 0

A quadratic equation ax 2 + bx + c = 0 has no real roots, if b 2 – 4ac < 0. That means, the quadratic equation contains imaginary roots.

14. The product of two consecutive positive integers is 360. To find the integers, this can be represented in the form of quadratic equation as

(a) x 2 + x + 360 = 0

(b) x 2 + x – 360 = 0

(d) x 2 – 2x – 360 = 0

Answer: (b) x 2 + x – 360 = 0

Let x and (x + 1) be the two consecutive integers.

According to the given,

x(x + 1) = 360

x 2 + x = 360

x 2 + x – 360

15. The equation which has the sum of its roots as 3 is

(a) 2x 2 – 3x + 6 = 0

(b) –x 2 + 3x – 3 = 0

(d) 3x 2 – 3x + 3 = 0

Answer: (b) –x 2 + 3x – 3 = 0

The sum of the roots of a quadratic equation ax 2 + bx + c = 0, a ≠ 0 is given by,

Coefficient of x / coefficient of x 2 = –(b/a)

Let us verify the options.

(a) 2x 2 – 3x + 6 = 0

Sum of the roots = – b/a = -(-3/2) = 3/2

(b) -x 2 + 3x – 3 = 0

Sum of the roots = – b/a = -(3/-1) = 3

2x 2 – 3x + √2 = 0

Sum of the roots = – b/a = -(-3/3) = 1

16. The quadratic equation 2x 2 – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(d) more than 2 real roots

Answer: (c) no real roots

2x 2 – √5x + 1 = 0

Comparing with the standard form of a quadratic equation,

a = 2, b = -√5, c = 1

b 2 – 4ac = (-√5) 2 – 4(2)(1)

= 5 – 8

= -3 < 0

Therefore, the given equation has no real roots.

17. The equation (x + 1) 2 – 2(x + 1) = 0 has

(a) two real roots

(b) no real roots

(d) two equal roots

Answer: (a) two real roots

(x + 1) 2 – 2(x + 1) = 0

x 2 + 1 + 2x – 2x – 2 = 0

x 2 – 1 = 0

18. The quadratic formula to find the roots of a quadratic equation ax 2 + bx + c = 0 is given by

(a) [-b ± √(b 2 -ac)]/2a

(b) [-b ± √(b 2 -2ac)]/a

(d) [-b ± √(b 2 -4ac)]/2a

Answer: (d) [-b ± √(b 2 -4ac)]/2a

The quadratic formula to find the roots of a quadratic equation ax 2 + bx + c = 0 is given by [-b ± √(b 2 -4ac)]/2a.

19. The quadratic equation x 2 + 7x – 60 has

(a) two equal roots

(b) two real and unequal roots

(b) no real roots

Answer: (b) two real and unequal roots

x 2 + 7x – 60 = 0

Comparing with the standard form,

a = 1, b = 7, c = -60

b 2 – 4ac = (7) 2 – 4(1)(-60) = 49 + 240 = 289 > 0

Therefore, the given quadratic equation has two real and unequal roots.

20. The maximum number of roots for a quadratic equation is equal to

Answer: (b) 2

The maximum number of roots for a quadratic equation is equal to 2 since the degree of a quadratic equation is 2.

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CBSE Class 10 Maths Case Study Questions for Chapter 4 Quadratic
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Case Study Questions Class 10 Maths Quadratic Equations
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Class 10 Maths: Case Study Questions of Chapter 4 Quadratic Equations
Now represent the following situations in the form of a quadratic equation. The sum of squares of two consecutive integers is 650. (a) x 2 + 2x - 650 = 0 (b) 2x 2 +2x - 649 = 0. (c) x 2 - 2x - 650 = 0 (d) 2x 2 + 6x - 550 = 0. Show Answer. The sum of two numbers is 15 and the sum of their reciprocals is 3/10. (a) x 2 + 10x - 150 = 0.
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A quadratic equation can be defined as an equation of degree 2. This means that the highest exponent of the polynomial in it is 2. The standard form of a quadratic equation is ax 2 + bx + c = 0, where a, b, and c are real numbers and $a \neq 0$ Every quadratic equation has two roots depending on the nature of its discriminant, D = b2 - 4ac.Based on the above information, answer the following ...
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Quadratic Equations Case Study Questions (CSQ's) Practice Tests. Timed Tests. Select the number of questions for the test: Select the number of questions for the test: TopperLearning provides a complete collection of case studies for CBSE Class 10 Maths Quadratic Equations chapter. Improve your understanding of biological concepts and develop ...
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NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations. A 1-mark question was asked from Chapter 4 Quadratic Equations in the year 2018. However, in the year 2017, a total of 13 marks were asked from the topic Quadratic Equations. Therefore, students need to have a thorough understanding of the topic.
Class 10 Maths Chapter 4 Quadratic Equations MCQs
MCQs for Class 10 Maths Chapter 4 (Quadratic Equations) are available online here at BYJU'S, along with answers. Also, get detailed explanations for each objective type of question along with the PDF of practice questions.
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Question 9. A quadratic equation can have at most _____ roots. Answer: two. Question 10. The quadratic equation x 2 - 3x + 5 = 0 has _____ roots. Answer: imaginary. Extra Questions for Class 10 Maths NCERT Solutions for Class 10 Maths

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