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MATHEMATICS TERM - II

Written as per the syllabus prescribed by the Central Board of Secondary Education.

CBSECLASS X

Printed at: Repro Knowledgecast Ltd., Mumbai

10441_11071_JUP

P.O. No. 34971

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

Salient Features

• Extensive coverage of the syllabus for Term - II in an effortless and easy to grasp format.

• In alignment with the latest paper pattern of Central Board of Secondary Education.

• Neat and labelled diagrams.

• Constructions drawn with accurate measurements.

• ‘Things to Remember’ highlights important facts.

• Variety of additional problems for practice.

• Questions from previous years board papers have been solved.

• Memory Maps at the end of each chapter to facilitate quick revision.

• Sample Test Paper at the end of each chapter designed for student’s Self Assessment.

• Model Question Papers in accordance with the latest paper pattern.

PREFACE In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you. “Class X: Mathematics” is a complete and thorough guide critically analyzed and extensively drafted to foster the student’s confidence. The book ensures extensive coverage of the syllabus for Term - II in an effortless and easy to grasp format.

The Topic-wise classified format for each chapter of this book helps the students in easy comprehension.

Each chapter includes the following features:

Theory of each mathematical concept is explained with appropriate references. NCERT exercises and Exemplar questions are covered with solutions. Problems based on NCERT exercises covering all the topics are provided with solutions. Practice problems based on NCERT exercises and MCQs help students to revise the topics thoroughly. One Mark Questions are provided for each chapter. Higher order thinking skills (HOTS) questions have been added for the student to gain insight on the various levels of theory-based questions. Value - Based Questions which emphasize on values have been included. Memory Map has been provided to give a quick overview of the chapter, helping the students in effective learning. Sample Test Paper at the end of each chapter helps to test the range of preparation of the students. ‘Things to Remember’ help the students’ gain knowledge required to understand different concepts.

Two Model Question Papers, designed as per CBSE Paper Pattern, are a unique tool to enable self-assessment for the students. Answers for previous years Board Questions have been included in this book. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.

Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! Yours faithfully,

Publisher

UNIT WISE WEIGHTAGE (Term ‐ II)

No. Units Marks

II Algebra (Contd.) 23

III Geometry (Contd.) 17

IV Trigonometry (Contd.) 08

V Probability 08

VI Coordinate Geometry 11

VII Mensuration 23

Total 90

SYLLABUS

Unit II: Algebra (Contd.) 3. Quadratic Equations: Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solution of the quadratic equations

(only real roots) by factorization, by completing the square and by using quadratic formula. Relationship discriminant and nature of roots.

Problems related to day to day activities to be incorporated. 4. Arithmetic Progressions: Motivation for studying Arithmetic Progression, Derivation of the nth term and sum of first n terms of

A.P. and their application in solving daily life problems. Unit III: Geometry (Contd.) 2. Circles: Tangents to a circle motivated by chords drawn from points coming closer and closer to the point. i. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of

contact. ii. (Prove) The lengths of tangents drawn from an external point to circle are equal. 3. Constructions: i. Division of a line segment in a given ratio (internally). ii. Tangent to a circle from a point outside it. iii. Construction of a triangle similar to a given triangle. Unit IV: Trigonometry (Contd.) 3. Heights and Distances: Simple problems on heights and distances. Problems should not involve more than two right triangles.

Angles of elevation / depression should be only 30, 45, 60. Unit V: Statistics and Probability 2. Probability: Classical definition of probability. Connection with probability as given in Class IX. Simple problems

on single events not using set notation. Unit VI: Coordinate Geometry 1. Lines (In two dimensions): Review the concepts of coordinate geometry done earlier including graphs of linear equations.

Awareness of geometraical representation of quadratic polynomials. Distance between two points and section formula (internal). Area of a triangle.

Unit VII: Mensuration 1. Areas Related to Circles: Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and

perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60, 90 and 120 only. Plane figures involving triangles, simple quadrilaterals and circle should be taken).

2. Surface areas and Volumes: i. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres,

hemispheres and right circular cylinders / cones. Frustum of a cone. ii. Problems involving converting one type of metallic solid into another and other mixed problems.

(Problems with combination of not more than two different solids be taken).

QUESTION PAPER

PATTER

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Contents

NCERT Textbook

Chapter No. Chapter Name Page No.

04 Quadratic Equations

1

05 Arithmetic Progressions

50

07 Coordinate Geometry

105

09 Some Applications of Trigonometry

153

10 Circles

184

11 Constructions

216

12 Areas Related to Circles

262

13 Surface Areas and Volumes

320

15 Probability

369

Model Question Paper – I

394

Model Question Paper – II

397

1

Chapter 04: Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0 where a, b, c are real numbers, a 0. (It is called standard form of a quadratic equation). Examples: 4x2 + 2x + 1 = 0, x2 + 4 = 0, 6 – 4x x2 = 0 are quadratic equations. Things to Remember 1. Check whether the following are quadratic

equations: i. (x + 1)2 = 2(x – 3) ii. x2 – 2x = (–2)(3 – x) iii. (x – 2) (x + 1) = (x – 1) (x + 3) iv. (x – 3) (2x + 1) = x (x + 5) v. (2x – 1) (x – 3) = (x + 5) (x – 1) vi. x2 + 3x + 1 = (x – 2)2 vii. (x + 2)3 = 2x (x2 – 1) viii. x3 – 4x2 – x + 1 = (x – 2)3 Solution: i. The given equation is (x + 1)2 = 2(x 3) x2 + 2x + 1 = 2x 6 x2 + 7 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. Alternate Method: LHS = (x + 1)2 = x2 + 2x + 1 RHS = 2(x 3) = 2x 6 (x + 1)2 = 2(x 3) can be written as x2 + 2x + 1 = 2x 6 x2 + 7 = 0 It is of the form ax2 + bx + c = 0 the given equation is a quadratic equation. ii. The given equation is x2 – 2x = (–2)(3 – x) x2 – 2x = 6 + 2x

x2 4x + 6 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. iii. The given equation is (x – 2) (x + 1) = (x – 1) (x + 3) x2 + x – 2x – 2 = x2 + 3x – x 3 x2 – x – 2 = x2 + 2x 3 3x – 1 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation.

(It is a linear equation). iv. The given equation is (x – 3) (2x + 1) = x (x + 5) 2x2 + x – 6x – 3 = x2 + 5x x2 – 10x – 3 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. v. The given equation is (2x – 1) (x – 3) = (x + 5) (x – 1) 2x2 – 6x – x + 3 = x2 – x + 5x – 5 2x2 – 7x + 3 = x2 + 4x – 5 x2 – 11x + 8 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. vi. The given equation is x2 + 3x + 1 = (x – 2)2 x2 + 3x + 1 = x2 – 4x + 4 7x – 3 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation. vii. The given equation is (x + 2)3 = 2x (x2 – 1) x3 + 6x2 + 12x + 8 = 2x3 – 2x x3 – 6x2 – 14x – 8 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation. viii. The given equation is x3 – 4x2 – x + 1 = (x – 2)3 x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8 2x2 – 13x + 9 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. 2. Represent the following situations in the

form of quadratic equations: i. The area of a rectangular plot is

528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

04. Quadratic Equations

NCERT Exercise 4.1

Things to Remember

Any equation of the form p(x) = 0 where p(x)is a quadratic polynomial is a quadraticequation.

ax2 + bx + c = 0, a 0 (when terms are

written in decreasing order of their degrees)is a standard form of the equation.

Quadratic Equations

2

Class X: Mathematics

2

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

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

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

Solution: i. Let the breadth of the rectangle be x m. the length of the rectangle = (2x + 1) m Area of rectangular plot = length breadth = (2x + 1) x m2 = (2x2 + x) m2 According to the given condition, 2x2 + x = 528 2x2 + x – 528 = 0 the required quadratic equation is

2x2 + x – 528 = 0 ii. Let the two consecutive integers be x and x + 1. Product of the two integers = x(x + 1) According to the given condition, x(x + 1) = 306 x2 + x = 306 x2 + x – 306 = 0 the required quadratic equation is x2 + x – 306 = 0 iii. Let Rohan’s present age be x years. Present age of Rohan’s mother = (x + 26) years After 3 years, Age of Rohan = (x + 3) years Age of Rohan’s mother = (x + 26 + 3) years = (x + 29) years According to the given condition, (x + 3) (x + 29) = 360 x2 + 29x + 3x + 87 = 360 x2 + 32x – 273 = 0 the required quadratic equation is x2 – 32x – 273 = 0 iv. Let the uniform speed of the train be x km/hr. Time taken by the train to cover a distance of

480 km = 480

xhrs ....

distance time =

speed

If the speed had been 8 km/h less, Then, speed of train = (x – 8) km/h Time taken by the train to cover a distance of

480 km = 480

8x hrs

According to the given condition,

480

8x – 480

x = 3

480 1 1

8x x

= 3

1 1

8x x

= 3

480

8

8

x x

x x

= 1

160

8

8x x = 1

160

x(x – 8) = (160) (8) x2 – 8x = 1280 x2 – 8x – 1280 = 0 the required quadratic equation is x2 – 8x – 1280 = 0 1. Which of the following are quadratic

equations? i. 3x2 7x + 5 = 0 ii. 3x3 – 8x2 + 8 = 5x + 5 iii. 10x = 4x2

iv. x + 1

x= 3x

v. (x – 4)2 + 1 = 5x – 3 vi. x(x + 2) + 9 = (x + 4) (x – 4) Solution: i. The given equation is 3x2 – 7x + 5 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. ii. The given equation is 3x3 – 8x2 + 8 = 5x + 5 3x3 – 8x2 – 5x + 3 = 0 It is not of the form ax2 + bx + c = 0 the given equation is not a quadratic equation. iii. The given equation is 10x = 4x2 4x2 – 10x = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. iv. The given equation is

x + 1

x= 3x

2 1x

x

= 3x

x2 + 1 = 3x2 2x2 – 1 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation.

Problems based on Exercise 4.1

3


v. The given equation is (x – 4)2 + 1 = 5x – 3 x2 – 8x + 16 + 1 = 5x – 3 x2 – 8x + 17 = 5x – 3 x2 – 13x + 20 = 0 It is of the form ax2 + bx + c = 0, a 0 the given equation is a quadratic equation. vi. The given equation is x(x + 2) + 9 = (x + 4) (x – 4) x2 + 2x + 9 = x2 – 16 2x + 25 = 0 It is not of the form ax2 + bx + c = 0, a 0 the given equation is not a quadratic equation. 2. In a cricket match Harbhajan took three

wickets less than twice the number of wickets taken by Zahir. The product of the number of wickets taken by these two is 20. Represent the above situation in the form of quadratic equation. [CBSE 2015]

Solution: Let the number of wickets taken by Zahir be x. Number of wickets taken by Harbhajan = 2x 3. According to the given condition, x(2x – 3) = 20 2x2 – 3x = 20 2x2 – 3x – 20 = 0 the required quadratic equation is 2x2 – 3x – 20 = 0. 3. The sum of two numbers is 27. We need to

find the numbers if sum of their reciprocal

is 3

4. Represent the above situation in the

form of quadratic equation. Solution: Let the first number be x second number = 27 – x.

Reciprocal of first number = 1

x

Reciprocal of second number = 1

27 x


1 1

27x x

= 3

4

27

27

x x

x x

= 3

4

27

27x x= 3

4

9

27x x= 1

4

36 = x(27 – x) 36 = 27x – x2 x2 – 27x + 36 = 0 the required quadratic equation is x2 – 27x + 36 = 0.

1. Check which of the following are quadratic

equations:

i. (x – 3)2 + 1 = 4x – 3

ii. x (2x + 4) = (x + 5) (x – 5)

iii. (x + 3)3 = x3 – 10

iv. x (4x + 5) = 4x2 + 5

v. (5x + 1) (4x + 2) = 20 (x – 1) (x – 2)

vi. 2x2 = 11

vii. 2

1x

x

= 12 7x

x

viii. 4x2 + 3 x + 9 = 3 x + 9 2. Represent the following situations in the

form of quadratic equations:

i. The product of two consecutive positive integers is 342. Represent this in the form of quadratic equation whose roots are these integers.

ii. The sum of the areas of two squares is 802 m2. The difference of their perimeter is 8 m. Represent a quadratic equation to find the sides of the two squares.

iii. Three consecutive even integers are such that the sum of the square of the first and the product of the other two is 484. Represent the quadratic equation to find out the integers.

iv. Sanika wishes to arrange three sticks together in the shape of a right triangle. The hypotenuse is 1 cm longer than the base and 8 cm longer than the altitude of the triangle. Form a quadratic equation to find the length of the smallest rod.

1. (i), (ii), (iii), (vi), (viii) are quadratic equations.

2. i. x2 + x – 342 = 0, where x is the smaller integer.

ii. x2 + 2x – 399 = 0, where x is side of smaller square.

iii. x2 + 3x – 238 = 0, where x is the smaller integer.

iv. x2 – 2x 15 = 0, where x is the length of smallest stick.

Practice Problems based on Exercise 4.1

Answers

4


4

1. Which of the following is a quadratic

equation? [NCERT Exemplar] (A) x2 + 2x + 1 = (4 – x)2 + 3

(B) 2x2 = (5 – x) 22

5x

(C) (k + 1) x2 + 3

2x = 7 where k = –1

(D) x3 – x2 = (x – 1)3 2. Which of the following is not a quadratic

equation? [NCERT Exemplar] (A) 2(x – 1)2 = 4x2 – 2x + 1 (B) 2x – x2 = x2 + 5

(C) 22 3x = 3x2 – 5x

(D) (x2 + 2x)2 = x4 + 3 + 4x2 3. Is 5 x2 + 6x + 3 = 0 a quadratic equation?

[CBSE 2012] (A) Yes (B) No (C) Can’t say (D) It is a linear equation 4. The standard form of quadratic equation is (A) ax3 + bx2 + cx + d = 0, a 0 (B) ax4 + cx + e = 0, a 0 (C) ax + b = 0, a 0 (D) ax2 + bx + c = 0, a 0 5. Which of the following is not a quadratic

equation? (A) (x – 5)2 + 1 = 4x – 3 (B) 3x (3x + 1) + 8 = 9(x + 2) (x – 2) (C) x (2x + 3) = x2 + 1 (D) (x + 2)3 = x3 – 4 6. Which of the following is a quadratic

equation? (A) (x + 1)2 = 5(x – 9) (B) (x – 6) (x + 7) = (x – 4) (x + 3) (C) x2 + 3x + 1 = (x – 2)2 (D) x3 – 1 = 0 In general, a real number is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a2 + b + c = 0. (where x = is solution of the quadratic equation). Factorisation Method: If the quadratic polynomial ax2 + bx + c can be expressed as the product of two linear factors, say (px + q) and (rx + s), where p, q, r, s R, such that p ≠ 0 and r ≠ 0,

then ax2 + bx + c = 0 can be written as

(px + q) (rx + s) = 0

px + q = 0 or rx + s = 0

q

p x or s

r x

Thus, q

p

and

s

r

are the roots of the quadratic

equation. Example:

x2 + 5x + 6 = 0

x2 + 3x + 2x + 6 = 0

x(x + 3) + 2(x + 3) = 0

(x + 3) (x + 2) = 0

x + 3 = 0 or x + 2 = 0

x = 3 or x = 2

x = –3 and x = –2 are two roots of the equation

x2 + 5x + 6 = 0. Note: The equation of the form (x – )2 = 0 has

repeated roots i.e. x = and x = 1. Find the roots of the following quadratic

equations by factorization:

i. x2 – 3x – 10 = 0

ii. 2x2 + x – 6 = 0

iii. 22 7 5 2 0 x x

iv. 2 12 + = 0

8x x

v. 100x2 – 20x + 1 = 0

Solution:

i. x2 – 3x – 10 = 0

x2 – 5x + 2x – 10 = 0

x (x – 5) + 2(x 5) = 0

(x – 5) (x + 2) = 0

x – 5 = 0 or x + 2 = 0

x = 5 or x = 2

the roots of the given equation are 2, 5.

Multiple Choice Questions

NCERT Exercise 4.2

Solution of a Quadratic Equation by Factorisation

Roots of the quadratic equationax2 + bx + c = 0 are the same as the zeroesof the quadratic polynomial ax2 + bx + c.

A quadratic equation can have atmost tworoots.

Things to Remember

5


ii. 2x2 + x – 6 = 0 2x2 + 4x – 3x – 6 = 0 2x (x + 2) – 3(x + 2) = 0 (2x – 3) (x + 2) = 0 2x – 3 = 0 or x + 2 = 0

x = 3

2 or x = 2

the roots of the given equation are 2, 3

2.

iii. 22 7 5 2 0 x x + 22 2 5 + 5 2 0 x x + x

2 2 5 2 0x x+ x+

2 5 2 0x+ x+

2 5 = 0x + or + 2 0x

5

=2

x or 2x =

the roots of the given equation are 5

2

, 2 .

iv. 2 1

28

x x = 0

16x2 – 8x + 1 = 0 16x2 – 4x – 4x + 1 = 0 4x (4x – 1) – 1(4x – 1) = x0 (4x – 1) (4x – 1) = 0 4x – 1 = 0 or 4x – 1 = 0

1 =

4x or 1

= 4

x


4, 1

4.

v. 100x2 – 20x + 1 = 0 100x2 – 10x – 10x + 1 = 0 10x (10x – 1) – 1(10x – 1) = 0 (10x – 1) (10x – 1) = 0 10x – 1 = 0 or 10x – 1 = 0

1=

10x or 1

=10

x

the roots of the given equation are 1 1,

10 10.

2. Represent the following situations

mathematically, and solve them. i. John and Jivanti together have 45 marbles.

Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

ii. A cottage industry produces a certain

number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the

total cost of production was ` 750. We would like to find out the number of toys produced on that day.

Solution: i. Let the number of marbles with John be x. Number of marbles with Jivanti = 45 – x. Number of marbles left with John when he

lost 5 marbles = x – 5 Number of marbles left with Jivanti when she

lost 5 marbles = 45 – x – 5 = 40 x According to the given condition, (x – 5) (40 – x) = 124 40x – x2 – 200 + 5x = 124 – x2 + 45x – 200 = 124 – x2 + 45x – 324 = 0 x2 – 45x + 324 = 0 x2 – 36x – 9x + 324 = 0 x(x – 36) – 9(x – 36) = 0 (x – 36) (x – 9) = 0 x – 36 = 0 or x – 9 = 0 x = 36 or x = 9 Number of marbles with Jivanti = 45 – x = 45 – 36 or 45 – 9 = 9 or 36 the number of marbles John and Jivanti had to

start with are either 36, 9 or 9, 36 respectively. ii. Let the number of toys produced in a day be x. Cost of production of each toy = ` (55 – x) The total cost of production on a particular

day = ` 750 According to the given condition, x(55 – x) = 750 55x – x2 = 750 x2 – 55x + 750 = 0 x2 – 30x – 25x + 750 = 0 x(x – 30) – 25(x – 30) = 0 (x – 30) (x – 25) = 0 x – 30 = 0 or x – 25 = 0 x = 30 or x = 25 the number of toys produced on that day are 25

or 30. 3. Find two numbers whose sum is 27 and

product is 182. Solution: Let one number be x. the other number = 27 – x According to the given condition, x(27 – x) = 182 27x – x2 = 182 x2 – 27x + 182 = 0 x2 – 14x – 13x + 182 = 0 x(x – 14) – 13(x – 14) = 0

6


6

(x – 14) (x – 13) = 0 x – 14 = 0 or x – 13 = 0 x = 14 or x = 13 other number = 27 x = 27 14 or 27 13 = 13 or 14 the two numbers are 13 and 14. 4. Find two consecutive positive integers, sum

of whose squares is 365. Solution: Let the two consecutive numbers be x and

x + 1. According to the given condition, x2 + (x + 1)2 = 365 x2 + x2 + 2x + 1 = 365 2x2 + 2x – 364 = 0 x2 + x – 182 = 0 x2 + 14x – 13x – 182 = 0 x(x + 14) – 13(x + 14) = 0 (x – 13) (x + 14) = 0 x – 13 = 0 or x + 14 = 0 x = 13 or x = 14 Since, x is positive integer x = 13 other number = x + 1 = 13 + 1 = 14 the two consecutive numbers are 13 and 14. 5. The altitude of a right triangle is 7 cm less

than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution: Let the base of the triangle be x cm. Altitude of the triangle = (x – 7) cm. By Pythagoras theorem, BC2 + AB2 = AC2 x2 + (x – 7)2 = (13)2 x2 + x2 – 14x + 49 = 169 2x2 – 14x – 120 = 0 x2 – 7x – 60 = 0 x2 – 12x + 5x – 60 = 0 x(x – 12) + 5(x – 12) = 0 (x – 12) (x + 5) = 0 x – 12 = 0 or x + 5 = 0 x = 12 or x = 5 But length of the side cannot be negative. x = 12 cm. Altitude of the triangle = x – 7 = 12 – 7 = 5 cm the length of other two sides are 5 cm and

12 cm.

6. 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 ` 90, find the number of articles produced and the cost of each article.

Solution: Let the number of articles produced be x. Cost of each article = ` (2x + 3) Total cost of production = x(2x + 3) According to the given condition, x(2x + 3) = 90 2x2 + 3x = 90 2x2 + 3x – 90 = 0 2x2 + 15x – 12x – 90 = 0 x(2x + 15) – 6(2x + 15) = 0 (x – 6) (2x + 15) = 0 x – 6 = 0 or 2x + 15 = 0

x = 6 or 15=

2

x

But the number of articles cannot be a negative fraction.

x = 6 Cost of each article = 2x + 3 = 2 6 + 3 = ` 15 the number of articles produced is 6 and the

cost of each article is ` 15. 1. If x =

1

2, is a solution of the quadratic

equation 3x2 + 2kx – 3 = 0, find the value of k. [CBSE 2015]

Solution: The given equation is 3x2 + 2kx – 3 = 0

Putting x = 1

2 , we get

2

1 13 2k 3 0

2 2

3k 3 0

4

3k 3

4

3 12k

4

9k

4

the value of k is 9

4

.

Problems based on Exercise 4.2 A

B C x

x – 713

7


2. Find the roots of the following quadratic equations by factorization:

i. x x2 3 +1 + 3 = 0 [CBSE 2015]

ii. 23 2 3x x = 0 [CBSE 2011, 2012]

iii. 24 3 + 5 2 3 = 0x x

[CBSE 2012, 2013] iv. abx2 + (b2 ac)x bc = 0 v. 9x2 – 6b2 x – (a4 – b4) = 0

[CBSE 2015] vi. (x – 1)2 – 5(x – 1) – 6 = 0 [CBSE 2015] vii. 2 9x + x = 13 Solution:

i. 2 3 1 + 3 0 x x

2 3 3 0 x x x +

3 1 3 0 x x x

3 1 0 x x

3 0 x or x – 1 = 0

= 3x or x = 1

the roots of the given equation are 3 , 1. ii. 23 2 3 0 x x

23 3 + 3 0 x x x

3 3 1 3 0 x x x

3 3 1 0x x

3 0 x or 3 +1 = 0x

= 3x or 1

=3

x


3,3

.

iii. 24 3 5 2 3 0 x x

24 3 8 3 2 3 0 x x x

4 3 + 2 3 3 + 2 0 x x x

4 3 3 + 2 0 x x

4 3 0 x or 3 + 2 = 0x

3=

2x or

2=

3x

the roots of the given equation are 3 2

,2 3

.

iv. abx2 + (b2 ac) x bc = 0 abx2 + b2x acx bc = 0 bx (ax + b) c (ax + b) = 0 (ax + b) (bx c) = 0

ax + b = 0 or bx c = 0

x = b

a

or x =

c

b

the roots of the given equation are b

a

,

c

b.

v. 9x2 – 6b2 x – (a4 – b4) = 0 9x2 – [3(a2 + b2) – 3(a2 – b2)] x

– (a2 + b2) (a2 – b2) = 0 9x2 – 3(a2 + b2) x + 3(a2 – b2) x

– (a2 + b2) (a2 – b2) = 0 3x [3x – (a2 + b2)] + (a2 – b2) [3x

– (a2 + b2)] = 0 [3x – (a2 + b2)] [3x + a2 – b2] = 0 3x – (a2 + b2) = 0 or 3x + a2 – b2 = 0 3x = a2 + b2 or 3x = b2 – a2

x = 2 2a b

3

or x =

2 2b a

3

the roots of the given equation are 2 2a b

3

,

2 2b a

3

.

vi. (x – 1)2 – 5(x – 1) – 6 = 0 x2 – 2x + 1 – 5x + 5 – 6 = 0 x2 – 7x = 0 x(x – 7) = 0 x = 0 or x – 7 = 0 x = 0 or x = 7 the roots of the given equation are 0, 7. vii. 2 9x + x = 13 2 9x = 13 – x 2x + 9 = (13 – x)2

….[Squaring both sides] 2x + 9 = 169 – 26x + x2 x2 – 28x + 160 = 0 x2 – 20x – 8x + 160 = 0 x(x 20) 8(x – 20) = 0 (x – 20) (x – 8) = 0 x 20 = 0 or x 8 = 0 x = 20 or x = 8 the roots of the given equation are 20, 8. 3. Find the roots of the following quadratic

equations:

i. 110 3, 0x x

x [CBSE 2012]

ii. 13 1 3

3

x x x

[CBSE 2011, 2012]

iii. 16 151 ; 0, 1

1x

x x

[CBSE 2014]

8


8

iv. 1 1 6

, 1, 51 5 7

x

x xfor x:

[CBSE 2010]

v.

2 1 3 9 30; 3,

3 2 3 3 2 3 2

x xx

x x x x

[CBSE 2012]

vi. 2 1 3 12 3 5; 3,

3 2 1 2

x xx

x x

[CBSE 2014]

vii. 1 1 1 1= + +

a + b + a bx x [CBSE 2012]

Solution:

i. 110 3 x

x

10x2 – 1 = 3x

10x2 – 3x – 1 = 0

10x2 5x + 2x – 1 = 0

5x (2x 1) + 1(2x 1) = 0

(5x + 1) (2x 1) = 0

5x + 1 = 0 or 2x 1 = 0

1

5

x or 1

2x

the roots of the given equation are

1 1,

5 2 .

ii. 13 1 3

3

x x x

x2 + 2x – 3 = 3x – 1

x2 – x – 2 = 0

x2 – 2x + x – 2 = 0

x(x – 2) + 1(x – 2) = 0

(x – 2) (x + 1) = 0

x – 2 = 0 or x + 1 = 0

x = 2 or x = –1

the roots of the given equation are 2, –1.

iii. 16 151

1

x x

16 15

1

x

x x

(16 – x) (x + 1) = 15x

16x + 16 – x2 – x = 15x

– x2 + 15x + 16 = 15x

x2 = 16

x = 16

x = 4

the roots of the given equation are 4, – 4.

iv. 1 1 6

1 5 7

x x

5 1 6

1 5 7

x x

x x

2

5 1 6

5 5 7

x x

x x x

2

6 6

4 5 7

x x

2

1 1

4 5 7x x

x2 + 4x 5 = 7 x2 + 4x 12 = 0 x2 + 6x 2x 12 = 0 x (x + 6) 2 (x + 6) = 0 (x + 6) (x 2) = 0 x + 6 = 0 or x 2 = 0 x = 6 or x = 2 the roots of the given equation are –6, 2.

v.

2 1 3 90

3 2 3 3 2 3

x x

x x x x

2x(2x + 3) + (x – 3) + (3x + 9) = 0 4x2 + 6x + x – 3 + 3x + 9 = 0 4x2 + 10x + 6 = 0 2x2 + 5x + 3 = 0 2x2 + 2x + 3x + 3 = 0 2x(x + 1) + 3(x + 1) = 0 (2x + 3) (x + 1) = 0 2x + 3 = 0 or x + 1 = 0

3

2 x or x = –1

But, x ≠ 3

2

x = –1 the root of the given equation is –1.

vi. 2 1 32 3 5

3 2 1

x x

x x

4 2 3 95

3 2 1

x x +

x x

4 2 2 1 3 9 35

3 2 1

x x x x

x x

(4x – 2) (2x – 1) – (3x + 9) (x + 3) = 5(x + 3) (2x – 1)

(8x2 – 4x – 4x + 2) – (3x2 + 9x + 9x + 27) = 5(2x2 – x + 6x – 3)

(8x2 – 8x + 2) – (3x2 +18x + 27) = 5(2x2 + 5x – 3)

5x2 – 26x – 25 = 10x2 + 25x – 15 5x2 + 51x + 10 = 0 5x2 + 50x + x + 10 = 0

9


5x(x + 10) + 1(x + 10) = 0 (5x + 1) (x + 10) = 0 5x + 1 = 0 or x + 10 = 0

1

5 x or x = –10


5

, –10.

vii. 1 1 1 1

a b a b

x x

1 1 1 1

a b a b

x x

a b a b

a b ab

x x

x x

a b a b

a b ab

x x

1 1

a b abx x

– ab = x(a + b + x) xa + xb + x2 + ab = 0 x2 + (a + b)x + ab = 0 x2 + ax + bx + ab = 0 x (x + a) + b (x + a) = 0 (x + a) (x + b) = 0 x + a = 0 or x + b = 0 x = – a or x = – b the roots of the given equation are –a, –b. 4. If (x2 + y2) (a2 + b2) = (ax + by)2. Prove that

a b

x y [CBSE 2014]

Solution: (x2 + y2) (a2 + b2) = (ax + by)2 x2a2 + x2b2 + y2a2 + y2b2

= a2x2 + b2y2 + 2abxy x2b2 + y2a2 – 2abxy = 0 (xb – ya)2 = 0 xb = ya

a b

x y

Hence proved. 5. The sum of ages (in years) of a son and his

father is 35 years and product of their ages is 150 years, find their ages.

[CBSE 2012, 2014] Solution: Let the age of son be x years Age of father = (35 – x) years According to the given condition, x(35 – x) = 150 35x – x2 = 150 x2 – 35x + 150 = 0

x2 – 30x – 5x + 150 = 0 x(x – 30) – 5(x – 30) = 0 (x – 30) (x – 5) = 0 x – 30 = 0 or x 5 = 0 x = 30 or x = 5 But age of son cannot be 30 years, because

that would mean father’s age = 5 years which is not possible.

x = 5 years Age of father = (35 – x) = 35 – 5 = 30 years the ages of son and his father are 5 years and

30 years respectively. 6. Find the two consecutive odd positive integers,

sum of whose square is 290. [CBSE 2012] Solution: Let the two consecutive odd positive integers

be x and x + 2. According to the given condition, x2 + (x + 2)2 = 290 x2 + x2 + 4x + 4 = 290 2x2 + 4x – 286 = 0 x2 + 2x – 143 = 0 x2 + 13x – 11x – 143 = 0 x(x + 13) – 11(x + 13) = 0 (x + 13) (x – 11) = 0 x + 13 = 0 or x – 11 = 0 x = –13 or x = 11 Since, x is positive integer, x = 11 other number = x + 2 = 11 + 2 = 13 the two consecutive odd positive integers are

11 and 13. 7. The difference of two numbers is 5 and the

difference of their reciprocals is 1

10. Find

the numbers. [CBSE 2012, 2014] Solution: Let one number be x Other number = x + 5 According to the given condition,

1 1 1

5 10

x x

5 1

5 10

x x

x x

5 1

5 10

x x

50 = x(x + 5) x2 + 5x – 50 = 0 x2 + 10x – 5x – 50 = 0 x(x + 10) – 5(x + 10) = 0 (x + 10) (x – 5) = 0

10


10

x + 10 = 0 or x – 5 = 0 x = – 10 or x = 5 Other number = x + 5 = 10 + 5 or 5 + 5 = 5 or 10 the two numbers are 5, 10 or 10, 5. 8. The total cost of a certain length of cloth is `

200. If the piece was 5 m longer and each metre of cloth costs ` 2 less, the cost of the piece would have remained unchanged. How longer is the piece and what is its original rate per metre? [CBSE 2015]

Solution: Let the length of the cloth be x m. Total cost of the cloth = ` 200

Cost per metre = ` 200

x

New length of the cloth = (x + 5) m

New cost per metre = ` 200

2 x

Total cost of the cloth = (x + 5) 2002

x


2005 2 200

xx

1000200 2 10 200 x

x

– 2x + 1000

x– 10 = 0

– 2x2 + 1000 – 10x = 0 x2 + 5x – 500 = 0 x2 + 25x – 20x – 500 = 0 x(x + 25) – 20(x + 25) = 0 (x + 25) (x – 20) = 0 x + 25 = 0 or x – 20 = 0 x = – 25 or x = 20 Length of cloth = 20 m

Original cost per metre = 200

x = ` 200

20= ` 10

the length of cloth is 20 m and its original rate is ` 10 per metre.

9. A shopkeeper buys a number of books for `1200. If he had bought 10 more books for the same amount, each book would have cost him `20 less. How many books did he buy? [CBSE 2012]

Solution: Let the number of books bought be x. Total cost of books = ` 1200

Cost of each book = ` 1200

x

New number of books = x + 10

Cost of each book = ` 1200

10x


1200 120020

10

x x

1200 10 120020

10

x x

x x

1200x + 12000 – 1200x = 20x(x + 10) 12000 = 20x2 + 200x 600 = x2 + 10 x x2 + 10x – 600 = 0 x2 + 30x – 20x – 600 = 0 x(x + 30) – 20(x + 30) = 0 (x + 30) (x – 20) = 0 x + 30 = 0 or x – 20 = 0 x = – 30 or x = 20 But, number of books cannot be negative. x = 20 the number of book bought is 20. 10. 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.

[CBSE 2012] Solution: Let B finish the work in x days. Time take by A to finish the work = (x 6) days

Part of work done by A in 1 day = 1

6x

Part of work done by B in 1 day = 1

x

Time taken by A and B to finish the work together = 4 days

Part of work done by A and B in 1 day = 1

4


1 1 1

6 4

x x

6 1

6 4

x x

x x

2 6 1

6 4

x

x x

4(2x – 6) = x(x – 6) 8x – 24 = x2 – 6x x2 – 14x + 24 = 0 x2 – 12x – 2x + 24 = 0 x(x – 12) – 2(x – 12) = 0 (x – 12) (x – 2) = 0 x – 12 = 0 or x – 2 = 0 x = 12 or x = 2 But x cannot be less than 6 x = 12 the time taken by B to finish the work is 12 days.

11


11. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original

fraction is 29

20. Find the original fraction.

[CBSE 2015] Solution: Let the denominator of the fraction be x Numerator of the fraction = x – 3

Original Fraction = 3x

x

New fraction = 3 2

2

x

x

1

2

x

x


3 1 29

2 20

x x

x x

2 3 1 29

2 20

x x x x

x x

2 2

2

3 2 6 29

2 20

x x x x x

x x

2

2

2 2 6 29

2 20

x x

x x

20(2x2 – 2x – 6) = 29(x2 + 2x) 40x2 – 40x – 120 = 29x2 + 58x 11x2 – 98x – 120 = 0 11x2 – 110x + 12x – 120 = 0 11x(x – 10) + 12(x – 10) = 0 (11x + 12) (x – 10) = 0 11x + 12 = 0 or x – 10 = 0

12=

11x or x = 10

x = 12

11

does not satisfy the given condition

x = 10

Original fraction = 3x

x

= 10 3

10

= 7

10

the original fraction is 7

10.

12. A motor boat, whose speed is 15 km/h in

still water, goes 30 km downstream and comes back in a total of 4 hours and 30 minutes. Determine the speed of streams.

[CBSE 2012, 2014] Solution: Let the speed of the stream = x km/h. Speed of motorboat in upstream = (15 – x) km/h Speed of motor boat in downstream = (15 + x) km/h

Downstream distance = Upstream distance = 30 km

Time required to go downstream = 30

15 x hrs

Time required to go upstream = 30

15 x hrs


30 30 14

15 15 2

x x

30 30 9

15 15 2

x x

30 15 30 15 9

15 15 2

x x

x x

2

450 30 450 30 9

225 2

x x

x

2

900 9

225 2x

2

100 1

225 2x

200 = 225 – x2 x2 = 25 x = 25 x = 5 But, speed cannot be negative. x = 5 the speed of the stream is 5 km/h. 13. A peacock is sitting on top of a pillar which

is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake, the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?

Solution: AB is the pillar. Peacock is at point A. Snake’s initial position is D. It is moving in

the direction of its hole which is at point B. Let the snake be caught at a distance x m from

the hole. Since the speed of the peacock and snake are

equal, they cover equal distance till point C. AC = DC = DB BC = (27 – x) m

A

9 m

B C Dx m

27 m

36


36

Since the roots of the equation are equal, b2 4ac = 0 (2p)2 4(p)(6) = 0 4p2 24p = 0 4p2 24p = 0 4p(p 6) = 0 4p = 0 or p 6 = 0 p = 0 or p = 6 But p ≠ 0 p = 6 the value of p is 6. v. (p 12)x2 + 2(p 12) x + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = p 12, b = 2(p 12), c = 2 Since the roots of the equation are equal, b2 4ac = 0 [2(p 12)]2 4(p 12) 2 = 0 4(p 12)2 8(p 12) = 0 4 (p 12) (p 12 2) = 0 (p 12) (p 14) = 0 p 12 = 0 or p 14 = 0 p = 12 or p = 14 But p ≠ 12 p = 14 the value of p is 14. 2. Find the value of p for which the quadratic

equation (p + 1) x2 6(p + 1)x + 3(p + 9) = 0 p 1, has equal roots hence find the roots of the equation. [CBSE 2015]

Solution: (p + 1)x2 6(p + 1)x + [3(p + 9)] = 0 Comparing with ax2 + bx + c = 0, we get a = p + 1, b = 6(p + 1), c = 3(p + 9) Since the roots of the equation are equal, b2 4ac = 0 [6(p + 1)]2 4(p + 1) [3(p + 9)] = 0 36 (p + 1)2 12(p + 1) (p + 9) = 0 12 (p + 1) [3(p + 1) (p + 9)] = 0 (p + 1) (3p + 3 p 9) = 0 (p + 1) (2p 6) = 0 p + 1 = 0 or 2p 6 = 0 p = 1 or p = 3 But p –1 p = 3 Substituting the value of p in the given

equation, we get (3 + 1)x2 6 (3 + 1)x + 3 (3 + 9) = 0 4x2 24x + 36 = 0 x2 6x + 9 = 0 (x 3)2 = 0 x = 3 or x = 3 the roots of the given equation are 3, 3.

3. If 2 is a root of the quadratic equation 3x2 + px 8 = 0 and the quadratic equation 4x2 2px + k = 0 has equal roots, find k.

[CBSE 2014]

Solution:

2 is a root of the equation 3x2 + px 8 = 0

3(2)2 + p(2) 8 = 0

12 + 2p 8 = 0

2p + 4 = 0

2p = 4

p = 2

Substituting the value of p in 4x2 2px + k = 0, we get

4x2 2(2) x + k = 0

4x2 + 4x + k = 0

Comparing with ax2 + bx + c = 0, we get

a = 4, b = 4, c = k

Since the roots of the equation are equal,

b2 4ac = 0

42 4(4)(k) = 0

16 16k = 0

16k = 16

k = 1

the value of k is 1. 4. Find the nature of the roots of the following

equation. If the real roots exists, find them:

3x2 4 3 x + 4 = 0 [CBSE 2012]

Solution:

3x2 4 3 x + 4 = 0


a = 3, b = 4 3 , c = 4

b2 4ac = 2

4 3 4(3)(4)

= 48 48

= 0

Since, b2 4ac = 0,

the roots are real and equal.

x = b b

and2a 2a

x =4 3 4 3

and2 3 2 3

x = 2 3 2 3

and3 3

the roots of the given equation are 2 3

3,

2 3

3.

37


5. Determine the positive value of ‘k’ for which the equation x2 + kx + 64 = 0 and x2 8x + k = 0 will both have real and equal roots. [CBSE 2014]

Solution: The first equation is x2 + kx + 64 = 0 Comparing with ax2 + bx + c = 0, we get Here a = 1, b = k, c = 64 Since the roots of the equation are equal, b2 4ac = 0 k2 4(1)(64) = 0 k2 = 256 k = 256 k = ± 16 ....(i) The second equation is x2 8x + k = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 8, c = k Since the roots of the equation are equal, b2 4ac = 0 (8)2 4(1)(k) = 0 64 4k = 0 4k = 64

k = 64

4

k = 16 ....(ii) From (i) and (ii), we get k = 16. the value of k is 16. 6. If the quadratic equation (1 + a2) b2x2 + 2abcx + c2 m2 = 0 in x has

equal roots prove that c2 = m2(1 + a2). [CBSE 2014, 2015]

Solution: (1 + a2) b2x2 + 2 abcx + (c2 m2) = 0 Comparing with Ax2 + Bx + C = 0, we get A = (1 + a2)b2, B = 2abc, C = c2 m2 Since the roots of the equation are equal, B2 4AC = 0 (2abc)2 4(1 + a2) b2(c2 m2) = 0 4a2b2c2 4(b2 + a2b2) (c2 m2) = 0 4a2b2c2 4[b2c2 b2m2 + a2b2c2 a2b2m2] = 0 4a2b2c2 4b2c2 + 4b2m2 4a2b2c2

+ 4a2b2m2 = 0 4b2 [a2m2 + m2 c2] = 0 a2m2 + m2 c2 = 0 c2 = a2m2 + m2 c2 = m2(1 + a2) Hence proved. 7. For what values of k will quadratic

equation (2k + 1)x2 + 2(k + 3)x + (k + 5) = 0 have real and equal roots? [CBSE 2012]

Solution: (2k + 1)x2 + 2(k + 3)x + (k + 5) = 0 Comparing with ax2 + bx + c = 0, we get a = 2k + 1, b = 2(k + 3), c = k + 5

Since the roots of the equation are equal, b2 4ac = 0 [2(k + 3)]2 4(2k + 1) (k + 5) = 0 4(k + 3)2 4(2k2 + 11k + 5) = 0 (k + 3)2 (2k2 + 11k + 5) = 0 k2 + 6k + 9 2k2 11k 5 = 0 k2 5k + 4 = 0 k2 + 5k 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 5, c = 4 By quadratic formula,

x = 2b b 4ac

2a

k =

25 5 4 1 4

2 1

k = 5 25 16

2

k = 5 41

2

k = 5 41

2

or k =

5 41

2

the given equation has real and equal roots

when the value of k is5 41 5 41

or .2 2

8. Find the nature of the roots of the

quadratic equation 13 3 x2 + 10x + 3 = 0 [CBSE 2012]

Solution: 13 3 x2 + 10x + 3 = 0 Comparing with ax2 + bx + c = 0, we get

a = 13 3 , b = 10, c = 3

b2 4ac = (10)2 4 13 3 3

= 100 156 = 56 < 0 Since b2 4ac < 0, the equation has no real roots. 9. Find whether the real roots of the quadratic

equation 2 x2 + 7x + 8 2 = 0 exist or not. [CBSE 2012]

Solution: The given equation is 2 x2 + 7x + 8 2 = 0 Comparing with ax2 + bx + c = 0, we get

a = 2 , b = 7, c = 8 2

b2 4ac = (7)2 4 2 8 2

= 49 64 = 15 < 0 Since, b2 4ac < 0, the equation has no real roots.

38


38

10. In px2 + 4 3 x + 3 = 0 find the value of p so that

i. the roots are real ii. the roots are not real iii. the roots are equal

[CBSE 2012] Solution: The given equation is px2 + 4 3 x + 3 = 0 Comparing with ax2 + bx + c = 0, we get

a = p, b = 4 3 , c = 3

b2 4ac = 2

4 3 4(p)(3) = 48 12p

i. If the roots are real, b2 4ac ≥ 0 48 12p ≥ 0 48 ≥ 12p 4 ≥ p p ≤ 4 ii. If the roots are not real, b2 4ac < 0 48 12p < 0 48 < 12p 4 < p p > 4 iii. If roots are equal, b2 4ac = 0 48 12p = 0 12p = 48 p = 4 11. Rina wishes to fit three rods together in a

shape of a right triangle. The hypotenuse is to be 1 cm longer than the base and 2 cm longer than the altitude. Is it possible to draw such type of a right triangle. If so what should be the lengths of the rods?

Solution: Let the length of the hypotenuse x cm. Length of the base = (x 1) cm Length of the altitude = (x 2) cm In right ABC, by Pythagoras theorem, AC2 = AB2 + BC2 x2 = (x 2)2 + (x 1)2 x2 = x2 4x + 4 + x2 2x + 1 x2 6x + 5 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 6, c = 5 b2 4ac = (6)2 4(1)(5) = 36 20 = 16 > 0 Since b2 4ac > 0, the roots are real and distinct.

Hence it is possible to can draw the triangle. The roots of the equation are given by

x = 2b b 4ac

2a

x =

6 16

2 1

x = 6 4

2

x = 6 4

2

or x =

6 4

2

x = 10

2 or x =

2

2

x = 5 or x = 1 But x ≠ 1 x = 5 cm AB = x 2 = 5 2 = 3 cm BC = x 1 = 5 1 = 4 cm the lengths of the three rods are 5 cm, 3 cm

and 4 cm. 1. State whether the following quadratic

equations have two distinct real roots. Justify your answer.

i. x2 – 3x + 4 = 0 ii. 2x2 + x – 1 = 0

iii. 2x2 – 6x +9

2 = 0

iv. 3x2 – 4x + 1 = 0 v. (x + 4)2 – 8x = 0

vi. (x – 2 )2– 2(x + 1) = 0

vii. 2 x2 3

2x +

1

2= 0

viii. x (1 – x) – 2 = 0 ix. (x – 1)(x + 2) + 2 = 0 x. (x + 1)(x – 2) + x = 0 Solution: i. No Justification: The given equation is x2 3x + 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 3, c = 4 b2 4ac = (3)2 4(1)(4) = 9 16 = 7 < 0 Since b2 4ac < 0, the roots of the given equation are not real.

NCERT Exemplar

x

(x 1)

(x 2)

A

B C

39


ii. Yes Justification: The given equation is 2x2 + x 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = 1, c = 1 b2 4ac = (1)2 4(2)(1) = 1 + 8 = 9 > 0 Since b2 4ac > 0, the roots of the given equation are real and

distinct. iii. No Justification:

The given equation is 2x2 6x + 9

2 = 0

4x2 12x + 9 = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = 12, c = 9 b2 4ac = (12)2 4(4)(9) = 144 144 = 0 Since, b2 4ac = 0 the roots of the given equation are real and

equal. iv. Yes Justification: The given equation is 3x2 4x + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 3, b = 4, c = 1 b2 4ac = (4)2 4(3)(1) = 16 12 = 4 > 0 Since, b2 4ac > 0 the roots of the given equation are real and

distinct. v. No Justification: The given equation is (x + 4)2 8x = 0 x2 + 8x + 16 8x = 0 x2 + 16 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 0, c = 16 b2 4ac = (0)2 4(1)(16) = 64 < 0 Since b2 4ac < 0, the roots of the given equation are not real. vi. Yes Justification: The given equation is (x 2 )2 2(x + 1) = 0

x2 2 2 x + 2 2x 2 = 0

x2 2 2 x 2x = 0

x2 (2 2 + 2)x = 0


a = 1, b = 2 2 2 , c = 0

b2 4ac = [(2 2 + 2)]2 4(1)(0)

= 8 + 8 2 + 4 0

= 12 + 8 2 > 0 Since b2 4ac > 0, the roots of the given equation are real and

distinct. vii. Yes Justification:

The given equation is 2 x2 3

2 x +

1

2 = 0

2x2 3x + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = 3, c = 1 b2 4ac = ( 3)2 4(2)(1) = 9 8 = 1 > 0 Since b2 4ac > 0, the roots of the given equation are real and

distinct. viii. No Justification: The given equation is x(1 x) 2 = 0 x x2 2 = 0 x2 x + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 1, c = 2 b2 4ac = (1)2 4(1)(2) = 1 8 = 7 < 0 Since b2 4ac < 0, the roots of the given equation are not real. ix. Yes Justification: The given equation is (x 1)(x + 2) + 2 = 0 x2 + 2x x 2 + 2 = 0 x2 + x = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 1, c = 0 b2 4ac = 12 4(1)(0) = 1 > 0 Since b2 4ac > 0, the roots of the given equation are real and

distinct. x. Yes Justification: The given equation is (x + 1) (x 2) + x = 0 x2 x 2 + x = 0 x2 – 2 = 0

40


40

Comparing with ax2 + bx + c = 0, we get a = 1, b = 0, c = 2 b2 4ac = 02 4(1)(2) = 8 > 0 Since b2 4ac > 0, the roots of the given equation are real and

distinct. 2. Write whether the following statements are

true or false. Justify your answers. i. Every quadratic equation has exactly

one root. ii. Every quadratic equation has at least

one real root. iii. Every quadratic equation has at least

two roots. iv. Every quadratic equations has at

most two roots. v. If the coefficient of x2 and the

constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.

vi. If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

vii. If in a quadratic equation the coefficient of x is zero, then the quadratic equation has no real roots.

Solution: i. False. Justification: Every quadratic equation has two roots. ii. False. Justification: If the discriminant b2 4ac < 0, then the

quadratic equation has no real roots. iii. False. Justification: Every quadratic equation has only two roots. It

cannot have more than two roots. iv. True. Justification: A quadratic equation cannot have more than

two roots. v. True. Justification: If the values of a and c are of opposite sign,

then b2 4ac > 0. Hence the roots are real. vi. True. Justification: If the values of a and c are of same sign, and b

is 0, then b2 4ac < 0 Hence the roots are not real. vii. Not always.

Justification: Consider the quadratic equation ax2 + c = 0 Comparing with Ax2 + Bx + C = 0, we get A = a, B = 0, C = c B2 4AC = 4ac If a and c have same sign, then B2 4AC < 0 the roots are not real. If a and c have opposite signs, then

B2 4AC > 0 the roots are real and distinct. 3. A quadratic equation with integral

coefficient has integral roots. Justify your answer.

Solution: Not always. Justification: Consider x2 + x + 2 = 0 The equation has integral coefficients a = 1, b = 1, c = 2 b2 4ac = 12 4(1)(2) = 1 8 = 7 < 0 Since, b2 4ac < 0 the equation has no real roots. 4. Does there exist a quadratic equation whose

coefficients are rational but both of its roots are irrational? Justify your answer.

Solution: Yes Justification: Consider the equation x2 3x + 1 = 0 The equation has rational coefficients a = 1, b = 3, c = 1 Now, b2 4ac = (3)2 4(1) (1) = 9 4 = 5 > 0 the roots are given by

x =

3 5

2 1

= 3 5

2

the roots are irrational. 5. Does there exist a quadratic equation whose

coefficients are all distinct irrationals but both the roots are rationals? Why?

Solution Yes Justification: Consider the equation 5 x2 5 5 x + 6 5 =

0 25 3 5 2 5 6 5 0 x x x

5 3 2 5 3 0 x x x x

3 5 2 5 0 x x

41


x 3 = 0 or 5 2 5 0 x x = 3 or x = 2 the roots of the equation are 2 and 3, which are

rational. 6. Is 0.2 a root of the equation x2 – 0.4 = 0?

Justify. Solution: No Justification: The given equation is x2 0.4 = 0 put x = 0.2 in the given equation L.H.S. = (0.2)2 0.4 = 0.04 0.4 = 0.36 ≠ 0. 0.2 is not a solution of the given equation. 7. If b = 0, c < 0, is it true that the roots of

x2 + bx + c = 0 are numerically equal and opposite in sign? Justify.

Solution: Yes Justification: Consider the equation ax2 + bx + c = 0 Now, a = 1, b = 0, and c < 0 the equation becomes x2 c = 0 x2 = c x = c x = c the roots of the equation are numerically equal

and opposite in sign. 8. Does (x 1)2 + 2(x + 1) = 0 have real roots?

Justify your answer. Solution: No Justification: The given equation is (x 1)2 + 2(x + 1) = 0 x2 2x + 1 + 2x + 1 = 0 x2 + 1 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 0, c = 1 b2 4ac = 0 4(1)(1) = 4 < 0 the roots of the given equation are not real. 9. Find whether the following equations have

real roots. If real roots exist, find them. i. 8x2 + 2x – 3 = 0 ii. 2x2 + 3x + 2 = 0 iii. 5x2 2x – 10 = 0

iv.

1

2 3x+

1

5x= 1, x

3,5

2 [CBSE 2011]

v. x2 + 5 5x 70 = 0 Solution: i. 8x2 + 2x – 3 = 0 Comparing with ax2 + bx + c = 0, we get a = 8, b = 2, c = –3 b2 – 4ac = (2)2 4(8)(3) = 4 + 96 = 100 > 0 Since, b2 4ac > 0 the roots are real and distinct The roots are given by

x = 2b b 4ac

2a

x = 2 100

2(8)

x = 2 10

16

x = 2 10

16

or x = 2 10

16

x = 8

16 or x = 12

16

x = 1

2 or x = 3

4

the roots of given the equation are 1

2,

3

4

.

ii. 2x2 + 3x + 2 = 0 Comparing with ax2 + bx + c = 0 , we get a = –2, b = 3, c = 2 b2 – 4ac = (3)2 – 4(2) (2) = 9 + 16 = 25 > 0 Since, b2 4ac > 0, the roots are real and distinct The roots are given by

x = 2b b 4ac

2a

x = 3 25

2( 2)

x = 3 5

4

x = 3 5

4

or x = 3 5

4

x = 2

4 or x = 8

4

x = 1

2

or x = 2


2

, 2.

iii. 5x2 2x – 10 = 0 Comparing with ax2 + bx + c = 0, we get a = 5, b = –2, c = –10 b2 4ac = (–2)2 4(5) (–10) = 4 + 200 = 204 > 0

42


42

Since, b2 – 4ac > 0, the roots are real and distinct The roots are given by

x = 2b b 4ac

2a

x = 2 204

2(5)

x = 2 2 51

10

x = 1 51

5

x = 1 51

5

or x = 1 51

5

the roots of given the equation are 1 51

5 5 ,

1 51

5 5 .

iv. 1

2 3x +

1

5x = 1

5 2 3

(2 3)( 5)

x x

x x

= 1

2

3 8

2 13 15

x

x x

=1

2x2 – 13x + 15 = 3x – 8 2x2 – 16x + 23 = 0 Comparing with ax2 + bx + c = 0, we get a = 2, b = –16, c = 23 b2 – 4ac = (–16)2 – 4(2) (23) = 256 184 = 72 > 0 Since, b2 – 4ac > 0, the roots are real and distinct. The roots are given by

x = 2b b 4ac

2a

x =( 16) 72

2(2)

x = 16 6 2

4

x = 8 3 2

2

x = 8 3 2

2

or x =

8 3 2

2

x = 4 + 3 2

2or x = 4 –

3 2

2

the roots of the given equation are 4 +3 2

2,

4 – 3 2

2.

v. x2 + 5 5x 70 = 0 Comparing with ax2 + bx + c = 0, we get

a = 1, b = 5 5 , c = 70

b2 – 4ac = 2

5 5 4(1) (70)

= 125 + 280 = 405 > 0 Since, b2 4ac > 0, the roots are real and distinct The roots are given by

x = 2b b 4ac

2a

x =5 5 405

2(1)

x = 5 5 9 5

2

x = 5 5 9 5

2

or x =

5 5 9 5

2

x = 4 5

2or x =

14 5

2

x = 2 5 or x = 7 5

the roots of the given equation are 2 5 , 7 5. 1. Find the values of p for which the following

equation has equal roots. i. 9x2 + 8px + 16 = 0 ii. 4x2 + px + 3 = 0 [CBSE 2014]

iii. 2x2 + px + 9

2 = 0

vi. px(x 3) + 9 = 0 [CBSE 2014] v. (p + 4)x2 + ( p + 1) x + 1 = 0 vi. x2 + px + 16 = 0 2. Find value of k for which the following

equation has equal roots. i. x2 kx + (k 1) = 0 ii. kx2 2kx + 6 = 0 [CBSE 2012] iii. kx2 + 4x + 1 = 0 [CBSE 2012] iv. 9x2 3kx + k = 0 [CBSE 2014] v. (k 5)x2 + 2(k 5) x + 2 = 0 vi. x2 2x(1 + 3k) + 7(3+ 2k) = 0

[CBSE 2012] vii. (k 2)x2 + 2(2k 3)x + (5k 6) = 0

[CBSE 2015]

viii. kx (x 2 5 ) + 10 = 0 [CBSE 2013, 2014]

ix. x2 2(k + 1) x + k2 = 0 [CBSE 2013] x. x2 4kx + k = 0 [CBSE 2012]

Practice Problems based on Exercise 4.4

43


xi. k2x2 2 (2k 1) x + 4 = 0 xii. (k + 1) x2 2 (k 1) x + 1 = 0 xiii. (k 12) x2 + 2 (k 12) x + 2 = 0 xiv. (3k + 1) x2 + 2(k + 1)x + 1 = 0

[CBSE 2014] 3. If 5 is a root of the quadratic equation

2x2 + px 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots find the value of k.

[CBSE 2014, 2016] 4. If x = 2 is a root of the equation

3x2 + 7x + p = 0, find the value of k so that the roots of the equation x2 + k(4x + k 1) + p = 0 are equal.

[CBSE 2015] 5. If 1 is a root of the equation 3x2 + ax 2 = 0

and the quadratic equation a(x2 + 6x) b = 0 has equal roots, find the value of b.

[CBSE 2014] 6. If the root of the quadratic equation

(a b)x2 + (b c)x + (c a) = 0 a b are equal then prove that b + c = 2a.

[CBSE 2012, 2016] 7. If the roots of the equation

(a2 + b2)x2 2(ac + bd)x + (c2 + d2) = 0 are equal then prove that ad = bc.

[CBSE 2014] 8. Find the values of k for which the equation

(3k + 1)x2 + 2 (k + 1)x + 1 has equal roots. Also find the roots. [CBSE 2014]

9. Find the values of k for which the quadratic

equations (k + 4)x2 + (k + 1)x + 1 = 0 has equal roots. Also find roots.

[CBSE 2013, 2014] 10. If x = 4 is a root of the equation

x2 + 2x + 4p = 0, find the value of k for which the equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has equal roots. [CBSE 2015]

11. Find the nature of roots of the following

equations: i. x2 + 10x + 39 = 0

ii. 5x2 16 5 x + 4 = 0

iii. 4

5x2

2

7 x + 1 = 0

iv. 2x2 5x + 1 = 0 [CBSE 2011, 2012]

v. 8x2 22x 21 = 0

vi. 2

2 52 0

x x

vii. 4x2 20x + 25 = 0

12. The hypotenuse of right triangle is 3 5 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 15 cm. Is it possible to draw such a right triangle. If so, how long are the legs.

[CBSE 2012, 2014] 13. A pole has to be erected at a point on the

boundary of a circular park of diameter 15 m in such a way that difference from two diametrically opposite fixed gates P and Q on the boundary is 3 m. Is it possible to do so? If yes, at what distance from the two gates should the pole be erected?

1. i. 3, 3 ii. 4 3 , 4 3

iii. 6, 6 iv. 0, 4 v. 3, 5 vi. 8, 8 2. i. 2 ii. 6 iii. 4 iv. 0, 4

v. 7 vi. 2, 10

9

vii. 1, 3 viii. 2

ix. 1

2

x. 6

xi. 1

4 xii. 0, 3

xiii. 12, 14 xiv. 0, 1

3. 7

4

4. –1, 2

3

5. – 9

8. k = 0, 1. Roots are

(k 1) k k 1

3k 1

9. k = 5, 3. Roots are

(k 1) k 5 k 3

2 k 4

10. 2, 10

9

11. i. Not real ii. Real and distinct iii. Not real iv. Not real v. Real and distinct vi. Real and distinct vii. Real and equal 12. Yes. 3 cm, 6 cm 13. Yes. At 9 m and 12 m from the gates

Answers

44


44

1. The quadratic equation 2x2 – 5 x + 1 = 0 has [NCERT Exemplar] (A) two distinct real roots (B) two equal real roots (C) no real roots (D) more than 2 real roots 2. Which of the following equations has two

distinct real roots? [NCERT Exemplar]

(A) 2x2 – 3 2 x + 9

4 = 0

(B) x2 + x – 5 = 0

(C) x2 + 3x + 2 2 = 0 (D) 5x2 – 3x + 1 = 0 3. Which of the following equations has no real

roots? [NCERT Exemplar]

(A) x2 – 4x + 3 2 = 0

(B) x2 + 4x – 3 2 = 0

(C) x2 – 4x 3 2 = 0

(D) 3x2 + 4 3x + 4 = 0 4. Values of k for which the quadratic equation

2x2 – kx + k = 0 has equal roots is [NCERT Exemplar] (A) 0 only (B) 4 (C) 8 only (D) 0, 8 5. (x2 + 1)2 – x2 = 0 has [NCERT Exemplar] (A) four real roots (B) two real roots (C) no real roots (D) one real root 6. The value of k for which roots of quadratic

equation kx2 + 2x + 3 = 0 are equal is:

(A) 1

3 (B) 1

3

(C) 3 (D) – 3 7. The value of c for which the equation

ax2 + bx + c = 0 has equal roots is [CBSE 2012]

(A) 2b

a (B)

2b

4a

(C) 2a

b (D)

2a

4b

8. If the equation x2 – 4x + a = 0 has no real

roots, then: (A) a < 4 (B) a 4 (C) a < 2 (D) a > 4

9. Which of the following equations has the equal roots?

(A) x2 + 5x 500 = 0

(B) 2x2 – x + 1

8= 0

(C) 10x2 – 3x – 1 = 0 (D) 6x2 + 7x 10 = 0 10. Discriminant of the quadratic equation

5x2 + 3x – 7 = 0 is (A) 131 (B) 131 (C) 149 (D) –149

11. If the discriminant of the equation

6x2 – bx + 2 = 0 is 1, then the value of b is (A) 7 (B) 7

(C) 7 (D) 7 1. Check whether the given equation

(x – 3)2 + 9 = 2x2 is quadratic or not. Solution: (x – 3)2 + 9 = 2x2 x2 – 6x + 9 + 9 = 2x2 2x2 – x2 + 6x – 18 = 0 x2 + 6x – 18 = 0 which is of the from ax2 + bx + c = 0 and a 0 the given equation is quadratic. 2. If x = –2 is a solution of the quadratic

equation 2x2 + 3kx + 5 = 0, find the value of k.

Solution: The given equation is 2x2 + 3kx + 5 = 0 x = – 2 is root of the equation 2(–2)2 + 3k(–2) + 5 = 0 8 – 6k + 5 = 0 13 – 6k = 0 6k = 13

k 13

=6

the value of k is 13

6.

3. Find the roots of the quadratic equation 7x2 – 12x – 4 = 0. Solution: 7x2 – 12x – 4 = 0 7x2 – 14x + 2x – 4 = 0 7x(x – 2) + 2(x – 2) = 0 (x – 2)(7x + 2) = 0 x – 2 = 0 or 7x + 2 = 0

x = 2 or x = 2

7

the roots of the given equation are 2, 2

7

.

Multiple Choice Questions

One Mark Questions

45


4. Find the positive root of 24 + 48x = 8

Solution:

24 + 48x = 8

On squaring both sides, we get 4x2 + 48 = 64 4x2 = 64 48 4x2 = 16 x2 = 4 x = 2 the positive root of the given equation is 2. 5. If the roots of the quadratic equation

x2 + px + 25 = 0 are equal then find the value of p.

Solution: The given equation is x2 + px + 25 = 0 Since, the roots of the equation are equal, b2 – 4ac = 0 p2 – 4(1)(25) = 0 p2 – 100 = 0 p = 10 the value of p is 10 or 10. 6. Find the nature of roots for the quadratic

equation 5x2 – 7x + 3 = 0 Solution: 5x2 – 7x + 3 = 0 Comparing with ax2 + bx + c = 0, we get

a = 5, b = 7 , c = 3

b2 – 4ac = (– 7 )2 – 4(5)(3) = 7 – 60 = – 53 < 0 Since b2 – 4ac < 0, the roots of the given equation are not real. 1. For the quadratic equation given below, if

ps ≠ qr, then prove that the equation has no real roots.

x2(p2 + q2) + 2x (pr + qs) + (r2 + s2) = 0 Solution: The given equation is x2(p2 + q2) + 2x (pr + qs) + (r2 + s2) = 0 Comparing with ax2 + bx + c = 0, we get a = (p2 + q2), b = 2(pr + qs), c = (r2 + s2) b2 4ac = [2(pr + qs)]2 4(p2 + q2)(r2 + s2) = 4[(pr + qs)2

(p2r2 + p2s2 + q2r2 + q2s2)] = 4 (p2r2 + 2prqs + q2s2 – p2r2

p2s2 q2r2 q2s2) = 4 (2prqs p2s2 q2r2) = 4 (p2s2 2prqs + q2r2) b2 4ac = 4 (ps qr)2 ....(i)

But ps ≠ qr .... [given] ps qr ≠ 0 (ps qr)2 > 0 ...(ii) b2 4ac < 0 ...[From (i) and (ii)] the roots of the given equation are not real. 2. From a station, two trains start at the same

time. One train moves in west direction and other in North direction. First train moves 5 km/hr faster than the second train. If after 2 hours, distance between the two trains is 50 km, find the average speed of each train. [CBSE 2012]

Solution: Suppose that the two trains start from point O

and move in West and North respectively. After 2 hours, the two trains reach points A

and B respectively very such that AB = 50 km. Let the average speed of train moving in West

direction be x km/hr. Average speed of train moving in North

direction = (x + 5) km/hr. Distance covered by first train is 2 hrs = OA = 2x km Distance covered by second train in 2 hrs = OB = 2 (x + 5) km In right triangle OAB, by Pythagoras theorem, AB2 = OA2 + OB2 502 = (2x)2 + [2(x + 5)]2 2500 = 4x2 + 4 (x2 + 10x + 25) 2500 = 4 [x2 + x2 + 10x + 25] 625 = 2x2 + 10x + 25 2x2 + 10x 600 = 0 x2 + 5x 300 = 0 x2 + 20x 15x 300 = 0 x(x + 20) 15(x + 20) = 0 (x + 20) (x 15) = 0 x + 20 = 0 or x 15 = 0 x = 20 or x = 15 But speed cannot be negative. x = 15 km/hr Speed of other train = x + 5 = 15 + 5 = 20 km/hr the average speeds of the two trains are

15 km/hr and 20 km/hr.

HOTS Questions

W A x km/hr O

N

E

B

(x + 5) km/hr

S

50 km

46


46

3. If a, b, c, p, q and r are real numbers such that 2(ac + pr) = bq, then prove that atleast one of the equations ax2 + bx + c = 0 and px2 + qx + r = 0 has real roots.

Solution: Given equations are ax2 + bx + c = 0, and px2 + qx + r = 0 Let the discriminants of the above two

equations be D1 and D2 respectively. D1 = b2 4ac D2 = q2 4pr Now, D1 + D2 = b2 4ac + q2 4pr = b2 + q2 4(ac + pr) = b2 + q2 2 2 (ac + pr) = b2 + q2 2bq

....[ 2(ac + pr) = bq]

= (b q)2 ≥ 0

....[ (b q)2 ≥ for all real values of b and q]

D1 + D2 ≥ 0 D1 ≥ 0 or D2 ≥ 0 Atleast one equation has real roots. 4. The denominator of a fraction is one more

than thrice the numerator. If the sum of the

fraction and its reciprocal is 11

314

find the

fraction. Solution: Let the numerator of the fraction be x. Denominator = 3x + 1

Fraction = 3 1

x

x +

Reciprocal of the fraction = 3 1x +

x


3 1

x

x + +

3 1x +

x =

113

14

(3 1) 53

(3 1 14

x x +

x x +

2 2

)

2+ 9 + 6 1

3

2

2

x x x +

x x =

53

14

14(10x2 + 6x +1) = 53(3x2 + x) 140x2 + 84x + 14 = 159x2 + 53x 19x2 – 31x – 14 = 0 Comparing with ax2 + bx +c = 0, we get a = 19 , b = –31, c = –14 By quadratic formula,

x = 2b b 4ac

2a

x = 2( 31) ± ( 31) 4(19)( 14)

2(19)

x = 31 ± 961 + 1064

2(19)

x = 31 ± 2025

38

x = 31 ± 45

38

x = 76

38 or x =

14

38

x = 2 or x = 7

19

But x ≠ 7

19

x = 2

Fraction = 3 1

x

x+ = 2 2

6 + 1 7

the fraction is 2

7.

5. Determine p so that the equation

x2 + 7px + 4 = 0 has no real roots. Solution: The given equation is x2 + 7px + 4 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 7p and c = 4 Now b2 – 4ac = (7p)2 – 4(1)(4) = 49p2 – 16 Since the quadratic equation has no real

roots, b2 4ac < 0 49p2 – 16 < 0 49p2 < 16

p2 < 16

49

p2 16

49< 0

p2 2

4

7

< 0

4

p +7

4p

7

< 0

Either 4p

7 > 0 and 4

p +7

< 0

i.e. 4p >

7 and 4

p7

which is impossible.

Or

4

p7

< 0 and 4

p +7

> 0

i.e. 4p <

7 and 4

p7

4

7 < p < 4

7

47


Value Based Questions 1. The sum of the lives of CFL Lamp and

ordinary lamp is 16 years. Twice the square of the life of a CFL Lamp exceeds the square of the life of ordinary lamp by 164. Is this situation possible? If so, determine the life of each lamp. Which type of lamp is eco-friendly and one should prefer? Which value is depicted in the problem?

Solution: Let the life of CFL lamp be x years. Life of ordinary lamp = (16 – x) years According to the given condition, 2x2 = (16 x)2 + 164 2x2 (16 x)2 = 164 2x2 (256 32x + x2) = 164 2x2 256 + 32x x2 164 = 0 x2 + 32x 420 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 32, c = 420 b2 4ac = (32)2 4(1)( 420) = 1024 + 1680 = 2704 > 0 Since b2 4ac > 0, the roots are real and distinct the given situation is possible. The roots are given by

x = 2b b 4ac

2a

x = 32 2704

2(1)

x = 32 52

2

x = 32 52

2

or x =

32 52

2

x = 20

2or x =

84

2

x = 10 or x = 42 But life of lamp cannot be negative. x = 10 years Life of CFL lamp = 10 years Life of ordinary lamp = 16 x = 16 10 = 6 years One should prefer CFL lamp since it is eco-

friendly and has longer life The value depicted in the problem is i. Environmental awareness.

2. If the price of petrol in increased by ` 5 per liter, a person will have to buy 2 litre less petrol for ` 300. Find the original price and the increased price of petrol.

i. Why the price of petrol increasing day by day?

ii. Why should we conserve fuels? Solution:

Let the price of petrol be x per litre.

Quantity of petrol for ` 300 = 300

xlitres

New price of petrol = ` (x + 5)

Quantity of petrol for ` 300 = 300

5x+ litres


300

x –

300

5x+= 2

1 1

3005x x+

= 2

5

( + 5)

x + x

x x

=

2

300

2

5 1

5 150

x x

x2 + 5x = 750 x2 + 5x – 750 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 5, c = 750 By quadratic formula,

x = 2b b 4ac

2a

x = 25 5 4(1)( 750)

2(1)

x = 5 25 3000

2

x =

5 3025

2

x = 5 55

2

x = 5 55

2

or x =

5 55

2

x = 50

2 or x =

60

2

x = 25 or x = 30 But price of petrol cannot be negative. x = 25 Original price of petrol = ` 25 per litre Increased price of petrol = 25 + 5 = ` 30 per liter. i. The price of petrol increasing day by

day because of increased consumption of petrol.

ii. We should conserve fuels as it takes millions of years to form fossil fuels, and the reserves are getting depleted day by day.

48


48

Memory Map

Nature of roots of ax2 + bx + c = 0, a 0

Quadratic Equations Second degree polynomial

( General form: ax2 + bx + c = 0, a 0)

Method to solve Quadratic equation

ax2 + bx + c = 0, a 0

D = b2 – 4ac

Factorisation Completing the Square Quadratic formula

x = 2b ± b 4ac

2a

Steps

1. x2 + b

ax + c

a= 0

2. x2 + b

ax = c

a

3. x2 + b

22a

x

+ 2

b

2a

=2

b

2a

– c

a

4. 2 2

2

b b 4ac

2a 4ax +

5. 2± b 4acb

2a 2ax +

6. x = 2b ± b 4ac

2a

ax2 + bx + c = 0 (px + q) (rx + s) = 0

x = q

p , s

r

i. if D > 0, roots are real anddistinct

ii. if D = 0, roots are real andequal

iii. if D < 0, roots are not real

49


Sample Test Paper

Total Marks: 20

1. Find the roots of the equation x2 + 4x + 4 = 0 [1] 2. Check whether the equation (4x – 1) (x – 3) = (2x – 4) (2x – 3) is a quadratic or not [1] 3. Find the nature of the roots of the given quadratic equation, and find the roots if they exist

(x 5)(x 3) = 5 [2]

4. Find the roots of the quadratic equation 5x2 – 2 10 x + 2 = 0 [2]

5. Solve for x : 4 6 10

;5 7 3

x x

x xx 5, 7 [3]

6. If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0 has equal roots. Find the value of k. [3] 7. A two digited number is such that the product of its digits is 28. When 27 is added to this number,

the digits interchange their places. Find the number. [4] 8. A bus travel at a certain average speed for a distance of 75 km and then travels a distance of 90 km

at an average speed of 10 km/h more than the first speed. If it take 3 hours to complete the total journey, find its first speed. [4]

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