Written as per the syllabus prescribed by the Central Board of Secondary Education. CBSE CLASS X MATHEMATICS

Written as per the syllabus prescribed by the Central Board of Secondary Education. CBSE CLASS X MATHEMATICS TERM - II Salient Features Etensive 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. Printed at: Repro Knowledgecast Ltd., Mumbai 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. 0_07_JUP P.O. No. 397

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 etensively drafted to foster the student s confidence. The book ensures etensive 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 eplained with appropriate references. NCERT eercises and Eemplar questions are covered with solutions. Problems based on NCERT eercises covering all the topics are provided with solutions. Practice problems based on NCERT eercises 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 : mail@targetpublications.org A book affects eternity; one can never tell where its influence stops. Yours faithfully, Publisher Best of luck to all the aspirants! UNIT WISE WEIGHTAGE (Term II) No. Units Marks II Algebra (Contd.) 3 III Geometry (Contd.) 7 IV Trigonometry (Contd.) 08 V Probability 08 VI Coordinate Geometry VII Mensuration 3 Total 90

Unit II: Algebra (Contd.) SYLLABUS 3. Quadratic Equations: Standard form of a quadratic equation a + b + 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.. Arithmetic Progressions: Motivation for studying Arithmetic Progression, Derivation of the n th term and sum of first n terms of A.P. and their application in solving daily life problems. Unit III: Geometry (Contd.). 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 eternal 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, 5, 60. Unit V: Statistics and Probability. 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. 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. 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 0 only. Plane figures involving triangles, simple quadrilaterals and circle should be taken).. 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 mied problems. (Problems with combination of not more than two different solids be taken).

QUESTION PAPER PATTERN Mathematics Code No. 0 Time : 3 Hour Marks : 90 No. Typology of Questions Very Short Answer (VSA) ( mark) Short Answer-I (SA) ( Marks) Short Answer-II (SA) (3 Marks) Long Answer (LA) ( Marks) Total Marks (%) Weightage. Remembering (Knowledge based Simple recall questions, to know specific facts, terms, concepts, principles or theories; Identify, define or recite, information) 3 3 6%. Understanding (Comprehension to be familiar with meaning and to understand conceptually, interpret, compare, contrast, eplain, paraphrase or interpret information) 3 6% 3. Application (Use abstract information in concrete situation, to apply knowledge to new situations; Use given content to interpret a situation, provide an eample or solve a problem) 3 %. 5. High Order Thinking skills (Analysis and Synthesis Classify, compare, contrast or differentiate between different pieces of information; Organize and/or integrate unique pieces of information from a variety of sources) Creating, Evaluation and Multi-Disciplinary (Generating new ideas, product or ways of viewing things. Appraise, judge, and/or justify the value or worth of a decision or outcome or to predict outcomes based on values) * 08 6% 8% TOTAL = 6 = 0 3 = 30 = 90 00% *One of the LA ( marks) will be to assess the values inherent in the tets.

NCERT Tetbook Chapter No. Contents Chapter Name Page No. 0 Quadratic Equations 05 Arithmetic Progressions 50 07 Coordinate Geometry 05 09 Some Applications of Trigonometry 53 0 Circles 8 Constructions 6 Areas Related to Circles 6 3 Surface Areas and Volumes 30 5 Probability 369 Model Question Paper I 39 Model Question Paper II 397

0. Quadratic Equations Chapter 0: Quadratic Equations Quadratic Equations A quadratic equation in the variable is an equation of the form a + b + c = 0 where a, b, c are real numbers, a 0. (It is called standard form of a quadratic equation). Eamples: + + = 0, + = 0, 6 = 0 are quadratic equations. a + b + c = 0, a 0 (when terms are written in decreasing order of their degrees) Things to is a Remember standard form of the equation.. Check whether the following are quadratic equations: i. ( + ) = ( 3) ii. = ( )(3 ) iii. ( ) ( + ) = ( ) ( + 3) iv. ( 3) ( + ) = ( + 5) v. ( ) ( 3) = ( + 5) ( ) vi. + 3 + = ( ) vii. ( + ) 3 = ( ) viii. 3 + = ( ) 3 i. The given equation is ( + ) = ( 3) + + = 6 + 7 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. Alternate Method: LHS = ( + ) = + + RHS = ( 3) = 6 ( + ) = ( 3) can be written as + + = 6 + 7 = 0 It is of the form a + b + c = 0 the given equation is a quadratic equation. ii. NCERT Eercise. Things to Remember Any equation of the form p() = 0 where p() is a quadratic polynomial is a quadratic equation. The given equation is = ( )(3 ) = 6 + iii. iv. + 6 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. The given equation is ( ) ( + ) = ( ) ( + 3) + = + 3 3 = + 3 3 = 0 It is not of the form a + b + c = 0, a 0 the given equation is not a quadratic equation. (It is a linear equation). The given equation is ( 3) ( + ) = ( + 5) + 6 3 = + 5 0 3 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. v. The given equation is ( ) ( 3) = ( + 5) ( ) 6 + 3 = + 5 5 7 + 3 = + 5 + 8 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. vi. vii. viii. The given equation is + 3 + = ( ) + 3 + = + 7 3 = 0 It is not of the form a + b + c = 0, a 0 the given equation is not a quadratic equation. The given equation is ( + ) 3 = ( ) 3 + 6 + + 8 = 3 3 6 8 = 0 It is not of the form a + b + c = 0, a 0 the given equation is not a quadratic equation. The given equation is 3 + = ( ) 3 3 + = 3 6 + 8 3 + 9 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation.. Represent the following situations in the form of quadratic equations: i. The area of a rectangular plot is 58 m. 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.

Class X: Mathematics ii. The product of two consecutive positive integers is 306. We need to find the integers. iii. Rohan s mother is 6 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 80 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. i. Let the breadth of the rectangle be m. the length of the rectangle = ( + ) m Area of rectangular plot = length breadth = ( + ) m = ( + ) m + = 58 + 58 = 0 the required quadratic equation is + 58 = 0 ii. Let the two consecutive integers be and +. Product of the two integers = ( + ) ( + ) = 306 + = 306 + 306 = 0 therequired quadratic equation is + 306 = 0 iii. iv. Let Rohan s present age be years. Present age of Rohan s mother = ( + 6) years After 3 years, Age of Rohan = ( + 3) years Age of Rohan s mother = ( + 6 + 3) years = ( + 9) years ( + 3) ( + 9) = 360 + 9 + 3 + 87 = 360 + 3 73 = 0 the required quadratic equation is 3 73 = 0 Let the uniform speed of the train be km/hr. Time taken by the train to cover a distance of 80 km = 80 distance hrs... time = speed If the speed had been 8 km/h less, Then, speed of train = ( 8) km/h Time taken by the train to cover a distance of 80 km = 80 hrs 8 80 8 80 = 3 80 = 3 8 8 = 3 80 8 = 8 60 8 8 = 60 ( 8) = (60) (8) 8 = 80 8 80 = 0 the required quadratic equation is 8 80 = 0 Problems based on Eercise.. Which of the following are quadratic equations? i. 3 7 + 5 = 0 ii. 3 3 8 + 8 = 5 + 5 iii. 0 = iv. + = 3 v. ( ) + = 5 3 vi. ( + ) + 9 = ( + ) ( ) i. The given equation is 3 7 + 5 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. ii. iii. iv. The given equation is 3 3 8 + 8 = 5 + 5 3 3 8 5 + 3 = 0 It is not of the form a + b + c = 0 the given equation is not a quadratic equation. The given equation is 0 = 0 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. The given equation is + = 3 = 3 + = 3 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation.

Chapter 0: Quadratic Equations v. The given equation is ( ) + = 5 3 8 + 6 + = 5 3 8 + 7 = 5 3 3 + 0 = 0 It is of the form a + b + c = 0, a 0 the given equation is a quadratic equation. vi. The given equation is ( + ) + 9 = ( + ) ( ) + + 9 = 6 + 5 = 0 It is not of the form a + b + c = 0, a 0 the given equation is not a quadratic equation.. 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 0. Represent the above situation in the form of quadratic equation. [CBSE 05] Let the number of wickets taken by Zahir be. Numberof wickets taken by Harbhajan = 3. ( 3) = 0 3 = 0 3 0 = 0 therequired quadratic equation is 3 0 = 0. 3. The sum of two numbers is 7. We need to find the numbers if sum of their reciprocal is 3. Represent the above situation in the form of quadratic equation. Let the first number be second number = 7. Reciprocal of first number = Reciprocal of second number = 7 = 3 7 7 = 3 7 7 = 3 7 9 = 7 36 = (7 ) 36 = 7 7 + 36 = 0 therequired quadratic equation is 7 + 36 = 0. Practice Problems based on Eercise.. Check which of the following are quadratic equations: i. ( 3) + = 3 ii. ( + ) = ( + 5) ( 5) iii. ( + 3) 3 = 3 0 iv. ( + 5) = + 5 v. (5 + ) ( + ) = 0 ( ) ( ) vi. = vii. 7 viii. + 3 + 9 = 3 + 9. Represent the following situations in the form of quadratic equations: i. The product of two consecutive positive integers is 3. Represent this in the form of quadratic equation whose roots are these integers. ii. The sum of the areas of two squares is 80 m. 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 8. 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 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. Answers. (i), (ii), (iii), (vi), (viii) are quadratic equations.. i. + 3 = 0, where is the smaller integer. ii. + 399 = 0, where is side of smaller square. iii. iv. + 3 38 = 0, where is the smaller integer. 5 = 0, where is the length of smallest stick. 3

Class X: Mathematics Multiple Choice Questions. Which of the following is a quadratic equation? [NCERT Eemplar] (A) + + = ( ) + 3 (B) = (5 ) 5 (C) (k + ) + 3 = 7 where k = (D) 3 = ( ) 3. Which of the following is not a quadratic equation? [NCERT Eemplar] (A) ( ) = + (B) = + 5 (C) 3 = 3 5 (D) ( + ) = + 3 + 3. Is 5 + 6 + 3 = 0 a quadratic equation? [CBSE 0] (A) Yes (B) No (C) Can t say (D) It is a linear equation. The standard form of quadratic equation is (A) a 3 + b + c + d = 0, a 0 (B) a + c + e = 0, a 0 (C) a + b = 0, a 0 (D) a + b + c = 0, a 0 5. Which of the following is not a quadratic equation? (A) ( 5) + = 3 (B) 3 (3 + ) + 8 = 9( + ) ( ) (C) ( + 3) = + (D) ( + ) 3 = 3 6. Which of the following is a quadratic equation? (A) ( + ) = 5( 9) (B) ( 6) ( + 7) = ( ) ( + 3) (C) + 3 + = ( ) (D) 3 = 0 Solution of a Quadratic Equation by Factorisation In general, a real number is called a root of the quadratic equation a + b + c = 0, a 0 if a + b + c = 0. (where = is solution of the quadratic equation). Factorisation Method: If the quadratic polynomial a + b + c can be epressed as the product of two linear factors, say (p + q) and (r + s), where p, q, r, s R, such that p 0 and r 0, then a + b + c = 0 can be written as (p + q) (r + s) = 0 p + q = 0 or r + s = 0 q or p q Thus, p equation. Eample: s r and s r + 5 + 6 = 0 + 3 + + 6 = 0 ( + 3) + ( + 3) = 0 ( + 3) ( + ) = 0 + 3 = 0 or + = 0 = 3 or = are the roots of the quadratic = 3 and = are two roots of the equation + 5 + 6 = 0. Note: The equation of the form ( ) = 0 has repeated roots i.e. = and = NCERT Eercise. Things to Remember Roots of the quadratic equation a + b + c = 0 are the same as the zeroes of the quadratic polynomial a + b + c. A quadratic equation can have atmost two roots.. Find the roots of the following quadratic equations by factorization: i. 3 0 = 0 ii. + 6 = 0 iii. 7 5 0 iv. + = 0 8 v. 00 0 + = 0 i. 3 0 = 0 5 + 0 = 0 ( 5) + ( 5) = 0 ( 5) ( + ) = 0 5 = 0 or + = 0 = 5 or = the roots of the given equation are, 5.

Chapter 0: Quadratic Equations ii. + 6 = 0 + 3 6 = 0 ( + ) 3( + ) = 0 ( 3) ( + ) = 0 3 = 0 or + = 0 = 3 or = the roots of the given equation are, 3. iii. 7+ 5 0 + 5 +5 0 + + + 5+ 0 5 0 + 5= 0 or + 0 5 = or = the roots of the given equation are 5,. iv. = 0 8 6 8 + = 0 6 + = 0 ( ) ( ) = 0 ( ) ( ) = 0 = 0 or = 0 = or = the roots of the given equation are,. v. 00 0 + = 0 00 0 0 + = 0 0 (0 ) (0 ) = 0 (0 ) (0 ) = 0 0 = 0 or 0 = 0 = or 0 = 0 the roots of the given equation are, 0 0.. Represent the following situations mathematically, and solve them. i. John and Jivanti together have 5 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is. 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. i. Let the number of marbles with John be. Number of marbles with Jivanti = 5. Number of marbles left with John when he lost 5 marbles = 5 Number of marbles left with Jivanti when she lost 5 marbles = 5 5 = 0 ( 5) (0 ) = 0 00 + 5 = + 5 00 = + 5 3 = 0 5 + 3 = 0 36 9 + 3 = 0 ( 36) 9( 36) = 0 ( 36) ( 9) = 0 36 = 0 or 9 = 0 = 36 or = 9 Number of marbles with Jivanti = 5 = 5 36 or 5 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. Cost of production of each toy = ` (55 ) The total cost of production on a particular day = ` 750 (55 ) = 750 55 = 750 55 + 750 = 0 30 5 + 750 = 0 ( 30) 5( 30) = 0 ( 30) ( 5) = 0 30 = 0 or 5 = 0 = 30 or = 5 the number of toys produced on that day are 5 or 30. 3. Find two numbers whose sum is 7 and product is 8. Let one number be. the other number = 7 (7 ) = 8 7 = 8 7 + 8 = 0 3 + 8 = 0 ( ) 3( ) = 0 5

Class X: Mathematics ( ) ( 3) = 0 = 0 or 3 = 0 = or = 3 other number = 7 = 7 or 7 3 = 3 or the two numbers are 3 and.. Find two consecutive positive integers, sum of whose squares is 365. Let the two consecutive numbers be and +. + ( + ) = 365 + + + = 365 + 36 = 0 + 8 = 0 + 3 8 = 0 ( + ) 3( + ) = 0 ( 3) ( + ) = 0 3 = 0 or + = 0 = 3 or = Since, is positive integer = 3 other number = + = 3 + = the two consecutive numbers are 3 and. 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 3 cm, find the other two sides. A 6 7 3 B C Let the base of the triangle be cm. Altitude of the triangle = ( 7) cm. By Pythagoras theorem, BC + AB = AC + ( 7) = (3) + + 9 = 69 0 = 0 7 60 = 0 + 5 60 = 0 ( ) + 5( ) = 0 ( ) ( + 5) = 0 = 0 or + 5 = 0 = or = 5 But length of the side cannot be negative. = cm. Altitude of the triangle = 7 = 7 = 5 cm the length of other two sides are 5 cm and 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. Let the number of articles produced be. Cost of each article = ` ( + 3) Total cost of production = ( + 3) ( + 3) = 90 + 3 = 90 + 3 90 = 0 + 5 90 = 0 ( + 5) 6( + 5) = 0 ( 6) ( + 5) = 0 6 = 0 or + 5 = 0 5 = 6 or = But the number of articles cannot be a negative fraction. = 6 Cost of each article = + 3 = 6 + 3 = ` 5 the number of articles produced is 6 and the cost of each article is ` 5. Problems based on Eercise.. If =, is a solution of the quadratic equation 3 + k 3 = 0, find the value of k. [CBSE 05] The given equation is 3 + k 3 = 0 Putting =, we get 3 k 30 3 k 3 0 3 k 3 3 k 9 k the value of k is 9.

Chapter 0: Quadratic Equations. Find the roots of the following quadratic equations by factorization: 3+ + 3=0 [CBSE 05] i. ii. 3 3 = 0 [CBSE 0, 0] iii. 3 +5 3 = 0 [CBSE 0, 03] iv. ab + (b ac) bc = 0 v. 9 6b (a b ) = 0 [CBSE 05] vi. ( ) 5( ) 6 = 0 [CBSE 05] vii. 9 + = 3 3 + 30 i. 3 + 3 0 3 3 0 3 0 30 or = 0 = 3 or = the roots of the given equation are 3,. ii. 3 3 0 3 3 + 3 0 3 3 0 3 3 3 0 3 0 or 3 += 0 = 3 or = 3 the roots of the given equation are 3,. 3 iii. 3 5 3 0 3 83 3 0 3 3 + 0 3 + 3 3 + 0 3 0 or 3 += 0 3 = or = 3 the roots of the given equation are 3,. 3 iv. ab + (b ac) bc = 0 ab + b ac bc = 0 b (a + b) c (a + b) = 0 (a + b) (b c) = 0 a + b = 0 or b c = 0 b c = or = a b the roots of the given equation are c b. b, a v. 9 6b (a b ) = 0 9 [3(a + b ) 3(a b )] (a + b ) (a b ) = 0 9 3(a + b ) + 3(a b ) (a + b ) (a b ) = 0 3 [3 (a + b )] + (a b ) [3 (a + b )] = 0 [3 (a + b )] [3 + a b ] = 0 3 (a + b ) = 0 or 3 + a b = 0 3 = a + b or 3 = b a a b b a = or = 3 3 the roots of the given equation are a b b a,. 3 3 vi. ( ) 5( ) 6 = 0 + 5 + 5 6 = 0 7 = 0 ( 7) = 0 = 0 or 7 = 0 = 0 or = 7 the roots of the given equation are 0, 7. vii. 9 + = 3 9 = 3 + 9 = (3 ).[Squaring both sides] + 9 = 69 6 + 8 + 60 = 0 0 8 + 60 = 0 ( 0) 8( 0) = 0 ( 0) ( 8) = 0 0 = 0 or 8 = 0 = 0 or = 8 the roots of the given equation are 0, 8. 3. Find the roots of the following quadratic equations: i. 0 3, 0 [CBSE 0] ii. 3 3 3 [CBSE 0, 0] 6 5 iii. ; 0, [CBSE 0] 7

Class X: Mathematics 8 iv. v. vi. vii. i. 6,, 5 5 7 for : [CBSE 00] 3 9 3 0; 3, 3 3 3 3 [CBSE 0] 3 3 5; 3, 3 [CBSE 0] = + + [CBSE 0] a+b+ a b 0 3 0 = 3 0 3 = 0 0 5 + = 0 5 ( ) + ( ) = 0 (5 + ) ( ) = 0 5 + = 0 or = 0 or 5 the roots of the given equation are,. 5 3 3 3 ii. + 3 = 3 = 0 + = 0 ( ) + ( ) = 0 ( ) ( + ) = 0 = 0 or + = 0 = or = the roots of the given equation are,. iii. 6 5 6 5 (6 ) ( + ) = 5 6 + 6 = 5 + 5 + 6 = 5 = 6 = 6 = the roots of the given equation are,. iv. 6 5 7 5 6 5 7 5 6 5 5 7 6 6 5 7 5 7 + 5 = 7 + = 0 + 6 = 0 ( + 6) ( + 6) = 0 ( + 6) ( ) = 0 + 6 = 0 or = 0 = 6 or = the roots of the given equation are 6,. v. 3 9 0 3 3 3 3 ( + 3) + ( 3) + (3 + 9) = 0 + 6 + 3 + 3 + 9 = 0 + 0 + 6 = 0 + 5 + 3 = 0 + + 3 + 3 = 0 ( + ) + 3( + ) = 0 ( + 3) ( + ) = 0 + 3 = 0 or + = 0 3 or = 3 But, = the root of the given equation is. vi. 3 3 5 3 3 + 9 5 3 39 3 3 5 ( ) ( ) (3 + 9) ( + 3) = 5( + 3) ( ) (8 + ) (3 + 9 + 9 + 7) = 5( + 6 3) (8 8 + ) (3 +8 + 7) = 5( + 5 3) 5 6 5 = 0 + 5 5 5 + 5 + 0 = 0 5 + 50 + + 0 = 0

Chapter 0: Quadratic Equations vii. 5( + 0) + ( + 0) = 0 (5 + ) ( + 0) = 0 5 + = 0 or + 0 = 0 or = 0 5 the roots of the given equation are, 0. 5 a b a b a b a b ab a b ab ab ab ab ab ab ab ab ab = (a + b + ) a + b + + ab = 0 + (a + b) + ab = 0 + a + b + ab = 0 ( + a) + b ( + a) = 0 ( + a) ( + b) = 0 + a = 0 or + b = 0 = a or = b the roots of the given equation are a, b.. If ( + y ) (a + b ) = (a + by). Prove that y [CBSE 0] a b ( + y ) (a + b ) = (a + by) a + b + y a + y b = a + b y + aby b + y a aby = 0 (b ya) = 0 b = ya y a b Hence proved. 5. The sum of ages (in years) of a son and his father is 35 years and product of their ages is 50 years, find their ages. [CBSE 0, 0] Let the age of son be years Age of father = (35 ) years (35 ) = 50 35 = 50 35 + 50 = 0 30 5 + 50 = 0 ( 30) 5( 30) = 0 ( 30) ( 5) = 0 30 = 0 or 5 = 0 = 30 or = 5 But age of son cannot be 30 years, because that would mean father s age = 5 years which is not possible. = 5 years Age of father = (35 ) = 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 90. [CBSE 0] Let the two consecutive odd positive integers be and +. + ( + ) = 90 + + + = 90 + 86 = 0 + 3 = 0 + 3 3 = 0 ( + 3) ( + 3) = 0 ( + 3) ( ) = 0 + 3 = 0 or = 0 = 3 or = Since, is positive integer, = other number = + = + = 3 the two consecutive odd positive integers are and 3. 7. The difference of two numbers is 5 and the difference of their reciprocals is 0. Find the numbers. [CBSE 0, 0] Let one number be Other number = + 5 5 0 5 5 0 5 5 0 50 = ( + 5) + 5 50 = 0 + 0 5 50 = 0 ( + 0) 5( + 0) = 0 ( + 0) ( 5) = 0 9

Class X: Mathematics + 0 = 0 or 5 = 0 = 0 or = 5 Other number = + 5 = 0 + 5 or 5 + 5 = 5 or 0 the two numbers are 5, 0 or 0, 5. 8. The total cost of a certain length of cloth is ` 00. If the piece was 5 m longer and each metre of cloth costs ` less, the cost of the piece would have remained unchanged. How longer is the piece and what is its original rate per metre? [CBSE 05] Let the length of the cloth be m. Total cost of the cloth = ` 00 Cost per metre = ` 00 New length of the cloth = ( + 5) m 00 New cost per metre = ` 00 Total cost of the cloth = ( + 5) 00 5 00 000 00 0 00 + 000 0 = 0 + 000 0 = 0 + 5 500 = 0 + 5 0 500 = 0 ( + 5) 0( + 5) = 0 ( + 5) ( 0) = 0 + 5 = 0 or 0 = 0 = 5 or = 0 Length of cloth = 0 m Original cost per metre = 00 = ` 00 0 = ` 0 the length of cloth is 0 m and its original rate is ` 0 per metre. 9. A shopkeeper buys a number of books for `00. If he had bought 0 more books for the same amount, each book would have cost him `0 less. How many books did he buy? [CBSE 0] Let the number of books bought be. Total cost of books = ` 00 Cost of each book = ` 00 New number of books = + 0 0 Cost of each book = ` 00 0 00 00 0 0 00 0 00 0 0 00 + 000 00 = 0( + 0) 000 = 0 + 00 600 = + 0 + 0 600 = 0 + 30 0 600 = 0 ( + 30) 0( + 30) = 0 ( + 30) ( 0) = 0 + 30 = 0 or 0 = 0 = 30 or = 0 But, number of books cannot be negative. = 0 the number of book bought is 0. 0. 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 days, find the time taken by B to finish the work. [CBSE 0] Let B finish the work in days. Time take by A to finish the work = ( 6) days Part of work done by A in day = 6 Part of work done by B in day = Time taken by A and B to finish the work together = days Part of work done by A and B in day = 6 6 6 6 6 ( 6) = ( 6) 8 = 6 + = 0 + = 0 ( ) ( ) = 0 ( ) ( ) = 0 = 0 or = 0 = or = But cannot be less than 6 = the time taken by B to finish the work is days.

Chapter 0: Quadratic Equations. The numerator of a fraction is 3 less than its denominator. If is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is 9. Find the original fraction. 0 [CBSE 05] Let the denominator of the fraction be Numerator of the fraction = 3 3 Original Fraction = 3 New fraction = 3 9 0 3 9 0 3 6 9 0 6 9 0 0( 6) = 9( + ) 0 0 0 = 9 + 58 98 0 = 0 0 + 0 = 0 ( 0) + ( 0) = 0 ( + ) ( 0) = 0 + = 0 or 0 = 0 = or = 0 = does not satisfy the given condition = 0 3 Original fraction = = 0 3 7 = 0 0 the original fraction is 7. 0. A motor boat, whose speed is 5 km/h in still water, goes 30 km downstream and comes back in a total of hours and 30 minutes. Determine the speed of streams. [CBSE 0, 0] Let the speed of the stream = km/h. Speed of motorboat in upstream = (5 ) km/h Speed of motor boat in downstream = (5 + ) km/h Downstream distance = Upstream distance = 30 km 30 Time required to go downstream = hrs 5 30 Time required to go upstream = hrs 5 30 30 5 5 30 30 9 5 5 305 305 9 5 5 50 3050 30 9 5 900 9 5 00 5 00 = 5 = 5 = 5 = 5 But, speed cannot be negative. = 5 the speed of the stream is 5 km/h. 3. A peacock is sitting on top of a pillar which is 9 m high. From a point 7 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? A 9 m B m C 7 m 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 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 = (7 ) m D

Class X: Mathematics Since the roots of the equation are equal, b ac = 0 (p) (p)(6) = 0 p p = 0 p p = 0 p(p 6) = 0 p = 0 or p 6 = 0 p = 0 or p = 6 But p 0 p = 6 the value of p is 6. v. (p ) + (p ) + = 0 Comparing with a + b + c = 0, we get a = p, b = (p ), c = Since the roots of the equation are equal, b ac = 0 [(p )] (p ) = 0 (p ) 8(p ) = 0 (p ) (p ) = 0 (p ) (p ) = 0 p = 0 or p = 0 p = or p = But p p = the value of p is.. Find the value of p for which the quadratic equation (p + ) 6(p + ) + 3(p + 9) = 0 p, has equal roots hence find the roots of the equation. [CBSE 05] (p + ) 6(p + ) + [3(p + 9)] = 0 Comparing with a + b + c = 0, we get a = p +, b = 6(p + ), c = 3(p + 9) Since the roots of the equation are equal, b ac = 0 [6(p + )] (p + ) [3(p + 9)] = 0 36 (p + ) (p + ) (p + 9) = 0 (p + ) [3(p + ) (p + 9)] = 0 (p + ) (3p + 3 p 9) = 0 (p + ) (p 6) = 0 p + = 0 or p 6 = 0 p = or p = 3 But p p = 3 Substituting the value of p in the given equation, we get (3 + ) 6 (3 + ) + 3 (3 + 9) = 0 + 36 = 0 6 + 9 = 0 ( 3) = 0 = 3 or = 3 the roots of the given equation are 3, 3. 36 3. If is a root of the quadratic equation 3 + p 8 = 0 and the quadratic equation p + k = 0 has equal roots, find k. [CBSE 0] is a root of the equation 3 + p 8 = 0 3() + p() 8 = 0 + p 8 = 0 p + = 0 p = p = Substituting the value of p in p + k = 0, we get () + k = 0 + + k = 0 Comparing with a + b + c = 0, we get a =, b =, c = k Since the roots of the equation are equal, b ac = 0 ()(k) = 0 6 6k = 0 6k = 6 k = the value of k is.. Find the nature of the roots of the following equation. If the real roots eists, find them: 3 3 + = 0 [CBSE 0] 3 3 + = 0 Comparing with a + b + c = 0, we get a = 3, b = 3, c = b ac = 3 (3)() = = 8 8 = 0 Since, b ac = 0, the roots are real and equal. b and a b a = 3 and 3 3 3 = 3 and 3 3 3 the roots of the given equation are 3 3, 3 3.

Chapter 0: Quadratic Equations 5. Determine the positive value of k for which the equation + k + 6 = 0 and 8 + k = 0 will both have real and equal roots. [CBSE 0] The first equation is + k + 6 = 0 Comparing with a + b + c = 0, we get Here a =, b = k, c = 6 Since the roots of the equation are equal, b ac = 0 k ()(6) = 0 k = 56 k = 56 k = ± 6...(i) The second equation is 8 + k = 0 Comparing with a + b + c = 0, we get a =, b = 8, c = k Since the roots of the equation are equal, b ac = 0 (8) ()(k) = 0 6 k = 0 k = 6 k = 6 k = 6 From (i) and (ii), we get k = 6. the value of k is 6....(ii) 6. If the quadratic equation ( + a ) b + abc + c m = 0 in has equal roots prove that c = m ( + a ). [CBSE 0, 05] ( + a ) b + abc + (c m ) = 0 Comparing with A + B + C = 0, we get A = ( + a )b, B = abc, C = c m Since the roots of the equation are equal, B AC = 0 (abc) ( + a ) b (c m ) = 0 a b c (b + a b ) (c m ) = 0 a b c [b c b m + a b c a b m ] = 0 a b c b c + b m a b c + a b m = 0 b [a m + m c ] = 0 a m + m c = 0 c = a m + m c = m ( + a ) Hence proved. 7. For what values of k will quadratic equation (k + ) + (k + 3) + (k + 5) = 0 have real and equal roots? [CBSE 0] (k + ) + (k + 3) + (k + 5) = 0 Comparing with a + b + c = 0, we get a = k +, b = (k + 3), c = k + 5 Since the roots of the equation are equal, b ac = 0 [(k + 3)] (k + ) (k + 5) = 0 (k + 3) (k + k + 5) = 0 (k + 3) (k + k + 5) = 0 k + 6k + 9 k k 5 = 0 k 5k + = 0 k + 5k = 0 Comparing with a + b + c = 0, we get a =, b = 5, c = By quadratic formula, b b ac = a 5 5 k = 5 56 k = 5 k = 5 5 k = or k = the given equation has real and equal roots 5 5 when the value of k is or. 8. Find the nature of the roots of the quadratic equation 3 3 + 0 + 3 = 0 [CBSE 0] 3 3 + 0 + 3 = 0 Comparing with a + b + c = 0, we get a = 3 3, b = 0, c = 3 b ac = (0) 3 3 3 = 00 56 = 56 < 0 Since b ac < 0, the equation has no real roots. 9. Find whether the real roots of the quadratic equation + 7 + 8 = 0 eist or not. [CBSE 0] The given equation is + 7 + 8 = 0 Comparing with a + b + c = 0, we get a =, b = 7, c = 8 b ac = (7) 8 = 9 6 = 5 < 0 Since, b ac < 0, the equation has no real roots. 37

Class X: Mathematics 0. In p + 3 + 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 0] The given equation is p + 3 + 3 = 0 Comparing with a + b + c = 0, we get a = p, b = 3, c = 3 b ac = 3 (p)(3) = 8 p i. If the roots are real, b ac 0 8 p 0 8 p p p ii. If the roots are not real, b ac < 0 8 p < 0 8 < p < p p > iii. If roots are equal, b ac = 0 8 p = 0 p = 8 p =. Rina wishes to fit three rods together in a shape of a right triangle. The hypotenuse is to be cm longer than the base and 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? Let the length of the hypotenuse cm. Length of the base = ( ) cm Length of the altitude = ( ) cm ( ) A B ( ) In right ABC, by Pythagoras theorem, AC = AB + BC = ( ) + ( ) = + + + 6 + 5 = 0 Comparing with a + b + c = 0, we get a =, b = 6, c = 5 b ac = (6) ()(5) = 36 0 = 6 > 0 Since b ac > 0, the roots are real and distinct. C Hence it is possible to can draw the triangle. The roots of the equation are given by = = b b ac a 6 6 = 6 = 6 6 or = = 0 or = = 5 or = But = 5 cm AB = = 5 = 3 cm BC = = 5 = cm the lengths of the three rods are 5 cm, 3 cm and cm. NCERT Eemplar. State whether the following quadratic equations have two distinct real roots. Justify your answer. i. 3 + = 0 ii. + = 0 iii. 6 + 9 = 0 iv. 3 + = 0 v. ( + ) 8 = 0 vi. ( ) ( + ) = 0 vii. 3 + = 0 viii. ( ) = 0 i. ( )( + ) + = 0. ( + )( ) + = 0 i. No The given equation is 3 + = 0 Comparing with a + b + c = 0, we get a =, b = 3, c = b ac = (3) ()() = 9 6 = 7 < 0 Since b ac < 0, the roots of the given equation are not real. 38

Chapter 0: Quadratic Equations ii. iii. Yes The given equation is + = 0 Comparing with a + b + c = 0, we get a =, b =, c = b ac = () ()() = + 8 = 9 > 0 Since b ac > 0, the roots of the given equation are real and distinct. No The given equation is 6 + 9 = 0 + 9 = 0 Comparing with a + b + c = 0, we get a =, b =, c = 9 b ac = () ()(9) = = 0 Since, b ac = 0 the roots of the given equation are real and equal. iv. Yes The given equation is 3 + = 0 Comparing with a + b + c = 0, we get a = 3, b =, c = b ac = () (3)() = 6 = > 0 Since, b ac > 0 the roots of the given equation are real and distinct. v. No The given equation is ( + ) 8 = 0 + 8 + 6 8 = 0 + 6 = 0 Comparing with a + b + c = 0, we get a =, b = 0, c = 6 b ac = (0) ()(6) = 6 < 0 Since b ac < 0, the roots of the given equation are not real. vi. Yes The given equation is ( ) ( + ) = 0 + = 0 = 0 ( + ) = 0 Comparing with a + b + c = 0, we get a =, b =, c = 0 b ac = [( + )] ()(0) = 8 + 8 + 0 = + 8 > 0 Since b ac > 0, the roots of the given equation are real and distinct. vii. Yes The given equation is 3 + = 0 3 + = 0 Comparing with a + b + c = 0, we get a =, b = 3, c = b ac = ( 3) ()() = 9 8 = > 0 Since b ac > 0, the roots of the given equation are real and distinct. viii. No The given equation is ( ) = 0 = 0 + = 0 Comparing with a + b + c = 0, we get a =, b =, c = b ac = () ()() = 8 = 7 < 0 Since b ac < 0, the roots of the given equation are not real. i. Yes The given equation is ( )( + ) + = 0 + + = 0 + = 0 Comparing with a + b + c = 0, we get a =, b =, c = 0 b ac = ()(0) = > 0 Since b ac > 0, the roots of the given equation are real and distinct.. Yes The given equation is ( + ) ( ) + = 0 + = 0 = 0 39

Class X: Mathematics Comparing with a + b + c = 0, we get a =, b = 0, c = b ac = 0 ()() = 8 > 0 Since b ac > 0, the roots of the given equation are real and distinct.. Write whether the following statements are true or false. Justify your answers. i. Every quadratic equation has eactly one root. ii. Every quadratic equation has at least one real root. iii. Every quadratic equation has at least iv. two roots. Every quadratic equations has at most two roots. v. If the coefficient of and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots. vi. If the coefficient of and the constant term have the same sign and if the coefficient of term is zero, then the quadratic equation has no real roots. vii. i. False. ii. iii. iv. If in a quadratic equation the coefficient of is zero, then the quadratic equation has no real roots. Every quadratic equation has two roots. False. If the discriminant b ac < 0, then the quadratic equation has no real roots. False. Every quadratic equation has only two roots. It cannot have more than two roots. True. A quadratic equation cannot have more than two roots. v. True. If the values of a and c are of opposite sign, then b ac > 0. Hence the roots are real. vi. vii. True. If the values of a and c are of same sign, and b is 0, then b ac < 0 Hence the roots are not real. Not always. Consider the quadratic equation a + c = 0 Comparing with A + B + C = 0, we get A = a, B = 0, C = c B AC = ac If a and c have same sign, then B AC < 0 the roots are not real. If a and c have opposite signs, then B AC > 0 the roots are real and distinct. 3. A quadratic equation with integral coefficient has integral roots. Justify your answer. Not always. Consider + + = 0 The equation has integral coefficients a =, b =, c = b ac = ()() = 8 = 7 < 0 Since, b ac < 0 the equation has no real roots.. Does there eist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer. Yes Consider the equation 3 + = 0 The equation has rational coefficients a =, b = 3, c = Now, b ac = (3) () () = 9 = 5 > 0 the roots are given by = 3 5 the roots are irrational. = 3 5 5. Does there eist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why? Solution Yes Consider the equation 5 5 5 + 6 5 = 0 5 3 5 5 6 5 0 5 3 5 3 0 3 5 5 0 0

Chapter 0: Quadratic Equations 3 = 0 or 5 5 0 = 3 or = the roots of the equation are and 3, which are rational. 6. Is 0. a root of the equation 0. = 0? Justify. No The given equation is 0. = 0 put = 0. in the given equation L.H.S. = (0.) 0. = 0.0 0. = 0.36 0. 0. is not a solution of the given equation. 7. If b = 0, c < 0, is it true that the roots of + b + c = 0 are numerically equal and opposite in sign? Justify. Yes Consider the equation a + b + c = 0 Now, a =, b = 0, and c < 0 the equation becomes c = 0 = c = c = c the roots of the equation are numerically equal and opposite in sign. 8. Does ( ) + ( + ) = 0 have real roots? Justify your answer. No The given equation is ( ) + ( + ) = 0 + + + = 0 + = 0 Comparing with a + b + c = 0, we get a =, b = 0, c = b ac = 0 ()() = < 0 the roots of the given equation are not real. 9. Find whether the following equations have real roots. If real roots eist, find them. i. 8 + 3 = 0 ii. + 3 + = 0 iii. 5 0 = 0 iv. + 3 5 =, 3,5 [CBSE 0] v. + 5 5 70 = 0 i. 8 + 3 = 0 Comparing with a + b + c = 0, we get a = 8, b =, c = 3 b ac = () (8)(3) = + 96 = 00 > 0 Since, b ac > 0 the roots are real and distinct The roots are given by = b b ac a 00 = (8) 0 = 6 0 = or = 6 = 8 6 or = 6 = or = 3 0 6 the roots of given the equation are, 3. ii. + 3 + = 0 Comparing with a + b + c = 0, we get a =, b = 3, c = b ac = (3) () () = 9 + 6 = 5 > 0 Since, b ac > 0, the roots are real and distinct The roots are given by = b b ac a 3 5 = ( ) 3 5 = 3 5 35 = or = = or = 8 = or = the roots of the given equation are,. iii. 5 0 = 0 Comparing with a + b + c = 0, we get a = 5, b =, c = 0 b ac = ( ) (5) ( 0) = + 00 = 0 > 0

Class X: Mathematics Since, b ac > 0, the roots are real and distinct The roots are given by = b b ac a 0 = (5) = 5 0 = 5 5 = 5 5 or = 5 5 the roots of given the equation are 5, 5 5 iv. 5. 5 5 3 + 5 = 53 = (3)( 5) 3 8 = 35 3 + 5 = 3 8 6 + 3 = 0 Comparing with a + b + c = 0, we get a =, b = 6, c = 3 b ac = ( 6) () (3) = 56 8 = 7 > 0 Since, b ac > 0, the roots are real and distinct. The roots are given by = b b ac a ( 6) 7 = () = 6 6 = 8 3 = 8 3 or = 8 3 = + 3 or = 3 the roots of the given equation are + 3, 3. v. + 5 5 70 = 0 Comparing with a + b + c = 0, we get a =, b = 5 5, c = 70 b ac = 5 5 () (70) = 5 + 80 = 05 > 0 Since, b ac > 0, the roots are real and distinct The roots are given by = b b ac a = 5 5 05 () = 5 59 5 5 59 5 = or = = 5 or = 5 5 59 5 = 5 or = 7 5 the roots of the given equation are 5, 7 5. Practice Problems based on Eercise.. Find the values of p for which the following equation has equal roots. i. 9 + 8p + 6 = 0 ii. + p + 3 = 0 [CBSE 0] iii. + p + 9 = 0 vi. p( 3) + 9 = 0 [CBSE 0] v. (p + ) + ( p + ) + = 0 vi. + p + 6 = 0. Find value of k for which the following equation has equal roots. i. k + (k ) = 0 ii. k k + 6 = 0 [CBSE 0] iii. k + + = 0 [CBSE 0] iv. 9 3k + k = 0 [CBSE 0] v. (k 5) + (k 5) + = 0 vi. ( + 3k) + 7(3+ k) = 0 [CBSE 0] vii. (k ) + (k 3) + (5k 6) = 0 [CBSE 05] viii. k ( 5 ) + 0 = 0 [CBSE 03, 0] i. (k + ) + k = 0 [CBSE 03]. k + k = 0 [CBSE 0]

Chapter 0: Quadratic Equations i. k (k ) + = 0 ii. (k + ) (k ) + = 0 iii. (k ) + (k ) + = 0 iv. (3k + ) + (k + ) + = 0 [CBSE 0] 3. If 5 is a root of the quadratic equation + p 5 = 0 and the quadratic equation p( + ) + k = 0 has equal roots find the value of k. [CBSE 0, 06]. If = is a root of the equation 3 + 7 + p = 0, find the value of k so that the roots of the equation + k( + k ) + p = 0 are equal. [CBSE 05] 5. If is a root of the equation 3 + a = 0 and the quadratic equation a( + 6) b = 0 has equal roots, find the value of b. [CBSE 0] 6. If the root of the quadratic equation (a b) + (b c) + (c a) = 0 a b are equal then prove that b + c = a. [CBSE 0, 06] 7. If the roots of the equation (a + b ) (ac + bd) + (c + d ) = 0 are equal then prove that ad = bc. [CBSE 0] 8. Find the values of k for which the equation (3k + ) + (k + ) + has equal roots. Also find the roots. [CBSE 0] 9. Find the values of k for which the quadratic equations (k + ) + (k + ) + = 0 has equal roots. Also find roots. [CBSE 03, 0] 0. If = is a root of the equation + + p = 0, find the value of k for which the equation + p( + 3k) + 7(3 + k) = 0 has equal roots. [CBSE 05]. Find the nature of roots of the following equations: i. + 0 + 39 = 0 ii. 5 6 5 + = 0 iii. 5 7 + = 0 iv. 5 + = 0 [CBSE 0, 0] v. 8 = 0 vi. 5 0 vii. 0 + 5 = 0. The hypotenuse of right triangle is 3 5 cm. If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be 5 cm. Is it possible to draw such a right triangle. If so, how long are the legs. [CBSE 0, 0] 3. A pole has to be erected at a point on the boundary of a circular park of diameter 5 m in such a way that difference from two diametrically opposite fied 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? Answers. i. 3, 3 ii. 3, 3 iii. 6, 6 iv. 0, v. 3, 5 vi. 8, 8. i. ii. 6 iii. iv. 0, v. 7 vi. 0, 9 vii., 3 viii. i.. 6 i. ii. 0, 3 iii., iv. 0, 3. 7., 3 5. 9 8. (k ) kk k = 0,. Roots are 3k 9. k = 5, 3. Roots are 0. 0, 9. 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. Yes. 3 cm, 6 cm (k ) k 5 k 3 k 3. Yes. At 9 m and m from the gates 3

Class X: Mathematics Multiple Choice Questions. The quadratic equation 5 + = 0 has [NCERT Eemplar] (A) two distinct real roots (B) two equal real roots (C) no real roots (D) more than real roots. Which of the following equations has two distinct real roots? [NCERT Eemplar] (A) 3 + 9 = 0 (B) + 5 = 0 (C) + 3 + = 0 (D) 5 3 + = 0 3. Which of the following equations has no real roots? [NCERT Eemplar] (A) + 3 = 0 (B) + 3 = 0 (C) 3 = 0 (D) 3 + 3 + = 0. Values of k for which the quadratic equation k + k = 0 has equal roots is [NCERT Eemplar] (A) 0 only (B) (C) 8 only (D) 0, 8 5. ( + ) = 0 has [NCERT Eemplar] (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 k + + 3 = 0 are equal is: (A) (B) 3 3 (C) 3 (D) 3 7. The value of c for which the equation a + b + c = 0 has equal roots is [CBSE 0] (A) (C) b a a b (B) (D) b a a b 8. If the equation + a = 0 has no real roots, then: (A) a < (B) a (C) a < (D) a > 9. Which of the following equations has the equal roots? (A) + 5 500 = 0 (B) + 8 = 0 (C) 0 3 = 0 (D) 6 + 7 0 = 0 0. Discriminant of the quadratic equation 5 + 3 7 = 0 is (A) 3 (B) 3 (C) 9 (D) 9. If the discriminant of the equation 6 b + = 0 is, then the value of b is (A) 7 (B) 7 (C) 7 (D) 7 One Mark Questions. Check whether the given equation ( 3) + 9 = is quadratic or not. ( 3) + 9 = 6 + 9 + 9 = + 6 8 = 0 + 6 8 = 0 which is of the from a + b + c = 0 and a 0 the given equation is quadratic.. If = is a solution of the quadratic equation + 3k + 5 = 0, find the value of k. The given equation is + 3k + 5 = 0 = is root of the equation ( ) + 3k( ) + 5 = 0 8 6k + 5 = 0 3 6k = 0 6k = 3 3 k = 6 the value of k is 3 6. 3. Find the roots of the quadratic equation 7 = 0. 7 = 0 7 + = 0 7( ) + ( ) = 0 ( )(7 + ) = 0 = 0 or 7 + = 0 = or = 7 the roots of the given equation are,. 7

Chapter 0: Quadratic Equations. Find the positive root of +8 = 8 + 8 = 8 On squaring both sides, we get + 8 = 6 = 6 8 = 6 = = the positive root of the given equation is. 5. If the roots of the quadratic equation + p + 5 = 0 are equal then find the value of p. The given equation is + p + 5 = 0 Since, the roots of the equation are equal, b ac = 0 p ()(5) = 0 p 00 = 0 p = 0 the value of p is 0 or 0. 6. Find the nature of roots for the quadratic equation 5 7 + 3 = 0 5 7 + 3 = 0 Comparing with a + b + c = 0, we get a = 5, b = 7, c = 3 b ac = ( 7 ) (5)(3) = 7 60 = 53 < 0 Since b ac < 0, the roots of the given equation are not real. HOTS Questions. For the quadratic equation given below, if ps qr, then prove that the equation has no real roots. (p + q ) + (pr + qs) + (r + s ) = 0 The given equation is (p + q ) + (pr + qs) + (r + s ) = 0 Comparing with a + b + c = 0, we get a = (p + q ), b = (pr + qs), c = (r + s ) b ac = [(pr + qs)] (p + q )(r + s ) = [(pr + qs) (p r + p s + q r + q s )] = (p r + prqs + q s p r p s q r q s ) = (prqs p s q r ) = (p s prqs + q r ) b ac = (ps qr)...(i) But ps qr... [given] ps qr 0 (ps qr) > 0...(ii) b ac < 0...[From (i) and (ii)] the roots of the given equation are not real.. 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 hours, distance between the two trains is 50 km, find the average speed of each train. [CBSE 0] N 50 km W A km/hr O S Suppose that the two trains start from point O and move in West and North respectively. After 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 km/hr. Average speed of train moving in North direction = ( + 5) km/hr. Distance covered by first train is hrs = OA = km Distance covered by second train in hrs = OB = ( + 5) km In right triangle OAB, by Pythagoras theorem, AB = OA + OB 50 = () + [( + 5)] 500 = + ( + 0 + 5) 500 = [ + + 0 + 5] 65 = + 0 + 5 + 0 600 = 0 + 5 300 = 0 + 0 5 300 = 0 ( + 0) 5( + 0) = 0 ( + 0) ( 5) = 0 + 0 = 0 or 5 = 0 = 0 or = 5 But speed cannot be negative. = 5 km/hr Speed of other train = + 5 = 5 + 5 = 0 km/hr the average speeds of the two trains are 5 km/hr and 0 km/hr. B ( + 5) km/hr E 5

Class X: Mathematics 3. If a, b, c, p, q and r are real numbers such that (ac + pr) = bq, then prove that atleast one of the equations a + b + c = 0 and p + q + r = 0 has real roots. Given equations are a + b + c = 0, and p + q + r = 0 Let the discriminants of the above two equations be D and D respectively. D = b ac D = q pr Now, D + D = b ac + q pr = b + q (ac + pr) = b + q (ac + pr) = b + q bq...[ (ac + pr) = bq] = (b q) 0...[ (b q) for all real values of b and q] 6 D + D 0 D 0 or D 0 Atleast one equation has real roots.. The denominator of a fraction is one more than thrice the numerator. If the sum of the fraction and its reciprocal is 3 find the fraction. Let the numerator of the fraction be. Denominator = 3 + Fraction = 3 + Reciprocal of the fraction = 3 + 3 + + 3 + = 3 (3+ ) 53 (3+ ) + 3 +9 + 6 = 53 (0 + 6 +) = 53(3 + ) 0 + 8 + = 59 + 53 9 3 = 0 Comparing with a + b +c = 0, we get a = 9, b = 3, c = By quadratic formula, = = b b ac a (9) ( 3) ± ( 3) (9)( ) = 3 ± 96 + 06 (9) 3 ± 05 = 38 3 ± 5 = 38 = 76 38 or = 38 7 = or = 9 7 But 9 = Fraction = 3 + = 6+ 7 the fraction is 7. 5. Determine p so that the equation + 7p + = 0 has no real roots. The given equation is + 7p + = 0 Comparing with a + b + c = 0, we get a =, b = 7p and c = Now b ac = (7p) ()() = 9p 6 Since the quadratic equation has no real roots, b ac < 0 9p 6 < 0 9p < 6 p < 6 9 p 6 9 < 0 p 7 < 0 p+ 7 p 7 < 0 Either p > 0 and p+ < 0 7 7 i.e. p> and p which is impossible. 7 7 Or p 7 < 0 and p+ 7 > 0 i.e. p< and p 7 7 < p < 7 7

Chapter 0: Quadratic Equations Value Based Questions. The sum of the lives of CFL Lamp and ordinary lamp is 6 years. Twice the square of the life of a CFL Lamp eceeds the square of the life of ordinary lamp by 6. 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? Let the life of CFL lamp be years. Life of ordinary lamp = (6 ) years = (6 ) + 6 (6 ) = 6 (56 3 + ) = 6 56 + 3 6 = 0 + 3 0 = 0 Comparing with a + b + c = 0, we get a =, b = 3, c = 0 b ac = (3) ()( 0) = 0 + 680 = 70 > 0 Since b ac > 0, the roots are real and distinct the given situation is possible. The roots are given by = = = = b b ac a 3 70 () 3 5 3 5 or = 3 5 = 0 or = 8 = 0 or = But life of lamp cannot be negative. = 0 years Life of CFL lamp = 0 years Life of ordinary lamp = 6 = 6 0 = 6 years One should prefer CFL lamp since it is ecofriendly and has longer life The value depicted in the problem is i. Environmental awareness.. If the price of petrol in increased by ` 5 per liter, a person will have to buy 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? Let the price of petrol be per litre. Quantity of petrol for ` 300 = 300 litres New price of petrol = ` ( + 5) Quantity of petrol for ` 300 = 300 + 5 litres 300 300 + 5 = 300 +5 = + 5 = ( +5) 300 5 5 50 + 5 = 750 + 5 750 = 0 Comparing with a + b + c = 0, we get a =, b = 5, c = 750 By quadratic formula, b b ac = a 5 5 ()( 750) = () 5 5 3000 = = 5 55 = 5 55 555 = or = = 50 or = 60 5 305 = 5 or = 30 But price of petrol cannot be negative. = 5 Original price of petrol = ` 5 per litre Increased price of petrol = 5 + 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. 7