QUADRATIC EQUATIONS QUADRATIC POLYNOMIAL

4 QUADRATIC EQUATIONS QUADRATIC POLYNOMIAL A polynomil of degree two is clled qudrtic polynomil. E.g. x + 4, x 5x + 6, x + x, x + x 6. A qudrtic polynomil cn hve t most three terms nmely, terms contining x, x nd constnt. The generl term of qudrtic polynomil in x is x + bx + c, where, b, c re rel numbers nd 0. Qudrtic polynomils re generlly denoted by f(x), g(x), h(x) etc. In the qudrtic polynomil f(x) = x + bx + c;, b, c re clled coefficients. If for x =, where is rel number, the vlue of qudrtic expression becomes zero, then is clled zero of x + bx + c. There will be two zeros for qudrtic polynomil. They re denoted by nd. Note: f(x) = x + bx + c is lso clled qudrtic expression. Illustrtion 1: Find whether (i). x = is zero of x 5x+ 6. (ii). x = -4, x = -1 re zeros of x + 5x + 4 (iii). x = - 1, x = - 1 4 re zeros of 6x + 5x + 1 (i). x = x 5x + 6 = 5() + 6 = 4 10 + 6 = 0 x = is zero of x 5x + 6. (ii). x = -4, x + 5x + 4 = (-4) + 5(-4) + 4 = 16 0 + 4 = 0 x = -4 is zero of x + 5x + 4 x = - 1 x + 5x + 4 = (-1) + 5(-1) + 4 = 1 5 + 4 = 0 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 8 x = -1 is zero of x + 5x + 4 x = - 4 nd x = -1 re zeros of x + 5x + 4. (iii). x = - 1, 1 6x + 5x + 1 = 6 x = - 1 4, 6x + 5x + 1 = 5 5 0 1 = 6 4 1 + 5 4 + 1 = 16 0. 1 + 5 + 1 = 6 1 9-5 + 1 x = - 1 is zero nd x = - 1 4 is not zero of 6x + 5x + 1. QUADRATIC EQUATION Zeros of qudrtic polynomil x + bx + c = 0 where, b, c re rel numbers nd 0 is found by solving the corresponding eqution x + bx + c = 0 clled qudrtic eqution. If the rel number nd re two zeros of the qudrtic polynomil x + bx + c, then nd re roots of the qudrtic eqution x + bx + c = 0. The generl form of qudrtic eqution is x + bx + c = 0 where, b, c re rel numbers nd 0. There will be two roots for qudrtic eqution nd cn be found by solving the eqution x + bx + c = 0. Roots re lso clled solutions of x + bx + c = 0. Exercise 1: (i) Check whether the following equtions re qudrtic or not: () 4x + 5x = 0 (b) x = 4 (c) 6x = 7x (d) (x 1) (x + 5) = 0 (e) (x + 1) (x + ) = 7(x + 1) (x + ) (f) x + x 4x + 5 = 0 (g) x - 1 x = x (x 0) (h) 8x 5 = (4x + ) (x + 1) (ii) Determine whether the given vlue of x is solution of the given eqution or not () x =, x 7x + 6 = 0 (b) x = - 5, x =, 6x + 11x 10 = 0 (c) x =, 5x 4x 1 = 0 (d) x = -, x =, x - x 4 = 0 (e) (x + 8 (x + 5) = 0, x = 8, x = 5. (iii) Find whether x = + b, x = b re roots of x x = b. SOLVING A QUADRATIC EQUATION BY FACTORIZATION Illustrtion : Solve 81x 64 = 0. 81x 64 = 0. (9x) (8) = 0 (9x + 8) (9 x 8) = 0 x = - 8 9 or x = 8 9 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 84 ASCENTPlus ASCENT Progrm & School x = - 8 8, 9 9 re solutions of 81x 64 = 0 Illustrtion : Solve 5x 7x 6 = 0. 5x 7x 6 = 0 5x 10x + x 6 = 0 5x (x ) + (x ) = 0 (x ) (5x + ) = 0 x = 0 or 5x + = 0 x = or x = - 5 x =, - 5 re solutions of 5x 7x - 6 = 0 Illustrtion 4: Solve x 1 x x 1 x (x 1, - ). x 1 x x 1 x 0 x 1 x x 1 x x 1 x x 1 x x 1 x x 1 x (x + 1) (x + ) + (x 1) (x ) (x 1) (x + ) = 0 x + x + + x x + (x + x ) = 0 - x x + 10 = 0 - (x + x 10) = 0 x + x 10 = 0 (x + 5) (x ) = 0 x = - 5, or x =. x = -5, re the solutions of the given eqution. 0 Exercise : (i) Solve: x + x + = 0 x 1 x 10 x, 4 x x 4 6 (iii) Solve: x 5 x (iv) Solve: bx + (b c)x bc = 0. (ii) Solve: SOLVING A QUADRATIC EQUATION BY THE METHOD OF COMPLETION OF A SQUARE: QUADRATIC FORMULA x + bx + c = 0 Dividing throughout by ( 0), we get x + b x + c = 0 1 By dding nd subtrcting the squre of coefficient of x, We get b b b c x. x 0 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 85 b b 4c x 4 If b 4c 0, then b b b 4c x b b 4c x x 4 b b 4c 4c is rel number Note: Here D= b 4c is clled discriminnt of the qudrtic eqution x + bx + c = 0 Cse I : Cse -II: If b 4c > 0, then the eqution hs two distinct rel roots, given by b b 4c =, = b b 4c If b 4c = 0 i.e. b = 4c then the eqution hs rel nd equl roots, given by = - b, = - b. In this cse eqution will hve repeted root. Cse -III: If b 4c < 0 then the eqution does not hve rel roots. Note: The qudrtic eqution x + bx + c = 0 will hve rel roots if D = b 4c 0. Illustrtion 5: Determine whether the following qudrtic eqution hve rel roots nd if they hve, find them (i). x + 11x 6 = 0 (ii). x - 6x + 9 = 0 (iii). x + x + 1 = 0 (iv). x 4x 9 = 0 (i). x + 11x 6 = 0 Compring this eqution with x + bx + c = 0 We hve =, b = 11, c = -6 Discriminnt D = b 4c = 11 4() (-6) = 11 + 48 = 169 D > 0, given eqution will hve two distinct rel roots sy, given by = = b b 4c b b 4c = = 11 169 11 1 1 4 4 11 169 11 1 6 4 4 the two roots re 1 nd 6. (ii). x 6x + 9 = 0 = 1, b = - 6, c = 9 D = b 4c = (-6) 4(1) (9) = 6 6 = 0. FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 86 ASCENTPlus ASCENT Progrm & School Eqution hs repeted root given by = (iii). x + x + 1 = 0 = 1, b = 1, c = 1 D = b 4c = 1 4(1) (1) = - < 0 The eqution does not hve rel roots. b = = (iv). x 4x 9 = 0 = 1, b = -4, c = -9 D = b 4c = (-4) 4(1) (-9) = 16 + 6 = 4 > 0 The eqution hs two roots given by 4 4 4 4, 4 4 4 4 x =, re the required solutions. 6 1 Illustrtion 6: Find the vlues of p for which the qudrtic eqution 6x + px + 6 = 0 hs rel roots. D = b 4c = p 4 6 6 = p 144 As the eqution hs rel roots D 0 p 1 0 (p + 1) (p 1) 0..(1) (1) holds good if (i). p + 1 0 nd p 1 0 p - 1, p 1 p 1. Or (ii). p + 1 0 nd p 1 0 p - 1 nd p 1 p - 1 Required vlued of p re p - 1 or p 1. Exercise : (i) Exmine which of the following qudrtic equtions hs rel root(s). If so, find the root(s). () 6x x = 0 (b) x x + 5 = 0 (c) x 4x + 4 = 0 (d) x + x 4 = 0 (e) x + 5 x 5 = 0 1 4 (f) x 1 x x 4 (x -1, -, -4) (g) y + (6 + 4)y + 8 = 0 (ii) Find the vlues of p for which the following equtions hve rel roots () px + x + 1 = 0 (b) x + 4x p = 0 (c) 5x + px + 5 = 0 (d) 4x + px + 4 = 0 Illustrtion 7: The sum of squres of two consecutive positive integers is 1. Find the integers. Let x be one of the positive integers. Then the other is x + 1 sum of squres of the integers = x + (x + 1) = 1 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 87 x + x + x + 1 1 = 0 x + x 0 = 0 (x + x 110) = 0 x + x 110 = 0 (x 10) (x 11) = 0 x = 10 or x = 11. consecutive positive integers re 10 nd 11. Illustrtion 8: One side of rectngle exceeds its other side by cm. If its re is 195 cm, determine the sides of the rectngle. Let one side be x cm Then other side will be (x + ) cm Are of rectngle = x (x + ) x (x + ) = 195 x + x 195 = 0 x + 15x 1x 195 = 0 x(x + 15) 1(x + 15) = 0 (x 1) (x + 15) = 0 x = 1 or x = -15. Since, side of the rectngle cnnot be negtive x = 1. sides of rectngle re x, x + = 1 cm, 15 cm. Illustrtion 9: The hypotenuse of right ngled tringle is 5 cm. The difference between the lengths of the other two sides of the tringle is 17 cm. Find the lengths of these sides. Let the length of the shorter side be x cm. Then, the length of the longer side = (x + 17) cm AB = x, BC = x + 17, CA = 5 A x 5 B x+17 C By Pythgors theorem AB + BC = 5 x + (x + 17) = 5 x + 4x 6 = 0 x + 17x 168 = 0 (x 7) (x + 4) = 0 x = 7, x = - 4 But side of tringle cnnot be negtive x = 7 Length of the shorter side x = 7 cm Let of the longer side = x + 17 = 7 + 17 = 4 cm. Exercise 4: (i) Find two consecutive positive even integers the sum of whose squre is 40. (ii) A two digit number is such tht product of its digits is 0. If the digits re interchnged the resulting number will exceed the previous by 9. Wht is the number? FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 88 ASCENTPlus ASCENT Progrm & School (iii) The sum of the ges of fther nd his son is 45 yers. Five yers go the product of their ges ws four times the fther s ge t tht time. Find their present ges. Reltion between roots nd coefficients of Qudrtic eqution: If, re roots of x + bx + c = 0 then b b 4c =, = b b 4c b b 4c b b 4c + = + = sum of roots = - b b b Coeffic ient of x Sum of roots = + = Coeffic ient of x b b 4c b b 4c = = b b 4c b b 4c 4c c 4 4 4 c cons tnt term Product of roots = = coefficient of x Some of Importnt Formule: 1. + = ( + ) -. ( + ) = ( + ) - ( + ). 4 + 4 = ( + ) - 4. ( - ) = ( + ) - 4 Tke x + bx + c = 0 Divide the whole eqution by ( since 0) b c x x 0 b c x x 0 x ( + )x + = 0 i.e. x (sum of the roots) x + Product of the roots = 0 The qudrtic eqution whose roots re, is given by x ( + )x + = 0. Illustrtion 10: If, re roots of x + bx + c = 0. Find the qudrtic eqution whose roots re (), (b) +, + (c), (d) 1, 1 4 4, re roots of x + bx + c = 0 + = - b, = c () We hve to find the eqution whose roots re, FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 89 for the required eqution b b sum of roots = + = ( + ) = product of roots = () () = 4 = 4 c 4c The eqution whose roots re, is x (sum )x + (product ) = 0 x + b x + 4c = 0 i.e. x + bx + 4c = 0 b b 6 (b) Sum = + + + = + + 6 = 6 product = ( + ) ( + ) = + + + 9 c b c b 9 = + ( + ) + 9 = 9 required eqution is x (sum) x + product = 0 x b 6 c b 9 - x 0 i.e. x (-b + 6)x + (c b + 9) = 0 b / b (c) Sum = 4 4 4 4 4 c Product = 4 4 16 Required eqution is x b 4 x + c 16 = 0 i.e. 16x + 4bx + c = 0 (d) Sum = 1 1 b / b c / c 1 1 1 1 product c / c required eqution is i.e. cx + bx + = 0 b x x 0 c c Exercise 5: (i) Form the qudrtic eqution whose roots re () -, - 5 (b) - 5, + 5 (ii) If, re roots of x + bx + c = 0 (c 0), find the vlues of () + (b) + 1 1 (c) (d) (iii) If, re roots of x + x + = 0, find the qudrtic eqution whose roots re 1 1 () -, - (b), (c), (d), (e) - 1, - 1 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 90 ASCENTPlus ASCENT Progrm & School FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 91 ANSWERS TO EXERCISES Exercise 1: (i). () yes (b) yes (c) yes (d) yes (e) yes (f) no (g) no (h) No (ii). () yes (b) yes (c). no (d) no (e) no (iii). yes Exercise : (i). x = -, - (ii). 5, 5 (iii). x =, (iv). c b, b Exercise : (i). () roots re, 1 (b) no roots (c) repeted root = 1 (e) roots re 5, 5 4 (g), (d) roots re 41 41, (f) 1, 1 (ii). () p 1 4 (b) p (c) p -10 or p 10 (d) p - 8 or p 8 Exercise 4: (i) 1, 14 (ii) 56 (iii) Fthers ge = 6 yers, son s ge = 9 yers Exercise 5: (i). () x + 7x + 10 = 0 (b) x - 4 x 1 = 0 (ii). () bc b (c) (d) c b (b) c c b c c (iii). () x x + = 0 (b) x + x + = 0 (c) x + x + 7 = 0 (d) 8x + x + = 0 (e) x + 5x + 6 = 0 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 9 ASCENTPlus ASCENT Progrm & School SOLVED PROBLEMS SECTION -A 1 1 1 x x 5 6 Problem 1: Solve x, 5 1 1 1 x x 5 6 x 5 x 1 x x 5 6 8 1 x x 5 6 48 = x + x 15 x + x 6 = 0 x = -9 or 7 Problem : Solve x 1 x 9 0. x x x x Obviously, given eqution is vlid only when x 0, x + 0 x x x x 9 0 x x x + 5x + = 0 (x + ) (x + 1) = 0 x + = 0 or x + 1 = 0 but x + 0 x = -1. Problem : Find the solutions set of x 4 5x + 6 = 0. x 4 5x + 6 = 0 (x ) 5x + 6 = 0 Let t = x t 5t + 6 = 0 (t ) (t ) = 0 t = or t = x = or x = x = or x = solution set x =, Problem 4: Solve x + -x = 0. x + -x = 0 x 1 + - = 0 let t = x x FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 9 t + 1 t - = 0 t t + 1 = 0 (t 1) = 0 t = 1 x = 1 x = 0 x = 0. Problem 5: Find two consectutive positive odd integers such tht sum of their squres is 90. Problem 6: If the first of the two required consecutive positive odd integers is x, then the second will be x + By the given condition x + (x + ) = 90 x + 4x 86 = 0 (x + 1) (x 11) = 0 since the required odd number is positive, we hve x = 11 required integers re 11, 1. Find the number hving two digits such tht it is 4 times the sum nd three times the product of its two digits. When x 0, x = 1 or x = or, or 9 nd y = 0 or 1 or or 9 Let the required number be 10x + y By given conditions 10x + y = 4(x + y) (1) 10x + y = xy () From (1) 6x = y x = y Substituting vlue of y in () 10x + x = x x 1x = 6x 6x 1x = 0 6x (x ) = 0 s x 0, x = Substituting x = in () 10 + y = y 0 + y = 6y 5y = 0 y = 4 Required number = 10x + y = 10 + 4 = 4. Problem 7: In cricket mtch Anil took one wicket less thn twice the number of wickets tken by Rvi. If the product of the number of wickets tken by them is 15, find the number of wickets tken by ech of them. Let the number of wickets tken by Anil be x nd tht by Rvi be y By given conditions x = y 1 (1) xy = 15 y = 15 x Substituting () in (1) 15 x = 1 x x + x 0 = 0 () FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 94 ASCENTPlus ASCENT Progrm & School (x 5) (x + 6) = 0 x = 5 or - 6 x cnnot be negtive x = 5 from () y = 15 5. Anil hs tken 5 wickets nd Rvi hs tken wickets. Problem 8: Find the vlues of m for which the eqution x -mx + 7m 1 = 0 hs equl roots. Compring the given eqution x mx + 7m 1 = 0 with x + bx + c = 0, = 1, b = -m, c = 7m 1 Discriminnt D = b 4c = (-m) 4(1) (7m 1) = 4m 8m + 48 If the roots re equl D = 0 4m 8m + 48 = 0 4(m 7m + 1) = 0 m 7m + 1 = 0 (m ) (m 4) = 0 m = or 4 Problem 9: Determine the eqution, sum of whose roots is 1 nd sum of their squres is 1. Let, be roots of the required qudrtic eqution + = 1, + = 1 (+ ) = + + 1 = 1 + = -1 = -6. required qudrtic eqution is x ( + )x + = 0 x x 6 = 0. Problem 10: If nd re roots of x + bx + c = 0, find the eqution whose roots re,. Let, be roots of x + bx + c = 0 + = - b, = c For the required eqution sum = b c = = b = c Product = 1 Required eqution is x (sum)x + product = 0 b c x - x 1 0 c cx - (b c) x+ c = 0. c c FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 95 SECTION -B Problem 1: If nd re roots of x + bx + c = 0 then prove tht b b Problem : Problem : + = - b, = c b b = b = = b c b b c b bc b c c If, re roots of x + px + q = 0 nd, re roots of x + rx + s = 0, prove tht ( - ) ( - ) ( - ) ( - ) = (q s) + (p r) (sp rq). + = - p, = q, + = - r, = s ( - ) ( - ) ( - ) ( - ) = [ - ( + ) + ] [ - ( + ) + ] = [ + r + s] [ + r+ s] = + r ( + ) + r + s( + ) + s + rs( + ) = q rpq + r q + s(p q) + s srp = (q s) + (p r) (sp rq) If the sum of the roots of the eqution x + bx + c = 0 is equl to sum of the squres of their reciprocls then show tht bc, c, b re in A.P. Let, be roots of the given eqution 1 1 Given tht + = + = + = b c b c b c b c FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 96 ASCENTPlus ASCENT Progrm & School c = b + c b b, c, bc re in A.P. Problem 4: If the roots of the eqution x + bx + c = 0 re in the rtio m : n, prove tht mnb = c(m + n) + = b, = c Problem 5: : = m : n m n m n (Applying Componendo nd Dividendo) m n m n m n m n 4 m n b / m n b c 4 m n b m n b 4c m n b (m n) = b (m + n) 4c( m + n) 4c( m + n) = b [(m + n) (m n) ] 4c(m + n) = b [4mn] mnb = c(m + n) If the roots of the eqution (b c)x + b(c )x + c( b) = 0 re equl. Show tht 1 1. c b Compring the given eqution with Ax + Bx + C = 0. A = (b c), B = b(c ), C = c( b) As the roots re equl D = B 4AC = 0 [b(c )] 4[(b c)] [c( b)] = 0 b + b c b c 4 bc + 4c 4bc + 4b c = 0 b + b c + b c 4 bc + 4c 4bc = 0 (b + bc c) = 0 b + bc = c dividing both sides by bc 1 1. c b Problem 6: Show tht the roots of the eqution (b c)x + (c )x + ( b) = 0 re rel. FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 97 Compring the given eqution with Ax + Bx + C = 0. A = (b c), B = (c ), C = ( b) Discriminnt D = B 4AC (c ) 4(b c) ( b) = c + c 4b + 4b + 4c 4bc = + 4b + c 4b 4bc + c = ( b + c) As D is perfect squre it is lwys positive D > 0 roots of given eqution re rel. Problem 7: Let, b, c be rel numbers with 0. Let, be the roots of the eqution x + bx + c = 0. Express the roots of x + bcx + c = 0 in terms of,.., re roots of x + bx + c = 0 then + = - b, = c Let 1, 1 be roots of x + bcx + c = 0 bc bc b c 1 + 1 = +.(1) c 1 1 = 1 1 = ( ) ( ).() From (1) nd (), 1 =, 1 = roots of x + bcx + c = 0 re nd. 1 + 1 = Problem 8: Find the condition tht one root of x + bx + c = 0 is squre of the other root. If is one root, then will be nother root + = b,.(1). = c = c.() (1) ( + ) b = + 6 + ( + b ) = c c c b b c c bc b (c + c bc ) = - b. c( + c) = bc b. FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 98 ASCENTPlus ASCENT Progrm & School Problem 9: If the roots of the eqution x px + q = 0 differ by unity then prove tht p = 4q + 1. Let, be the roots of x px + q = 0 + = p, = q given - = 1 ( - ) = 1 ( + ) - 4 = 1 p 4q = 1 p = 4q + 1. Problem 10: If, re the roots of x + bx + c = 0 ( 0) nd + k, + k re roots of Ax + Bx + C = 0 (A 0) for some constnt k, then prove tht b 4c B 4AC. A, re roots of x + bx + c = 0 b + =, = c + k, + k re the roots of Ax + Bx + C = 0 B ( + k) + ( + k) =, A ( + k) ( + k) = C A ( + k) - ( + k) = - Squring both sides, [( + k) - ( + k)] = ( - ) [( + k) + ( + k)] 4( + k) ( + k) = ( + ) - 4 B C b c 4 4 A A B 4AC b 4c A FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 99 ASSIGNMENT PROBLEMS SUBJECTIVE SECTION A 1. Solve 4x 4 x+ ( 4 b 4 ) = 0 by fctoriztion method.. Solve 9x 9( + b)x + ( + 5b + b ) = 0 by fctoriztion method.. Solve x x b b x b x b (x, x b) by fctoriztion method. 4. Using qudrtic formul, solve p x + (p q )x q = 0. 5. Find the vlue of k for which (k + 1)x + (k + )x + (k + 8) = 0 hs equl roots. 6. If the roots of (1 + m )x + mcx + (c ) = 0 hs equl roots prove tht c = (1 + m ). 7. If the roots of (c b)x ( bc) x + (b c) = 0 re equl then show tht either = 0 or + b + c = bc. 8. Find the vlue of k, if the roots of the eqution x + kx + 64 = 0 nd x 8x + k = 0 (k > 0) re equl. 9. A fst trin tkes one hour less thn slow trin for journey of 00 km. If the speed of the slow trin is 10 km/hr less thn tht of the fst trin, find the speed of the two trins. 10. Two numbers differ by, nd their products is 504. Find the numbers. FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 100 ASCENTPlus ASCENT Progrm & School SECTION B 1. If one root of qudrtic eqution x + bx + c = 0 is equl to nth power of the other root, show 1 1 n 1 n1 n tht n c c b 0.. Show tht the eqution ( + b )x + ( + b)x + 1 = 0 hs no rel roots.. If the rtio of roots of the qudrtic eqution x + px + q = 0 be equl rtio of roots of x + lx + m = 0. Prove tht p m = l q. 4. If the roots of the eqution (x ) (x b) k = 0 re c nd d, then prove tht the roots of (x c) (x d) + k = 0 re nd b. 5. Let p nd q be the roots of x x + A = 0 nd let r nd s be the roots of x 18x + B = 0. If p < q < r < s re in rithmetic progression, find the vlues of A nd B. 6. Show tht if the roots of ( + b )x + x(c + bd) + c + d = 0 re rel, they will be equl. 7. If, re the roots of the eqution x px + q = 0 nd 1, 1 re the roots of the eqution x 1 1 qx + p = 0. Form the qudrtic eqution whose roots re 1 nd 1 1 1 1. 1 8. Show tht the roots of x + mx + n = 0 re rel when m = k + n k. 9. If one root is equl to the squre of the other root of the eqution x + x k = 0, wht is the vlue of k? 10. Let p(x) = x + x + b be qudrtic polynomil in which nd b re integers. Given ny integer n, show tht there is n integer m such tht p(n) p(n + 1) = p(m). FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 101 OBJECTIVE Multiple Choice Questions: 1. If one root of qudrtic eqution x + kx 6 is then the other root is (A) -1 (B) (C) (D). The roots of 4x 4x + ( b ) = 0 re (A) + b, b (B) b, b (C) + b, b (D) b, b. The roots of x 1 + x 1 0 re ( 0) (A), 1 (B) -, - 1 (C), - 1 (D) -, 1 4. The vlue of k if (k + 1)x (k 1) x + 1 = 0 hs equl roots (A) 0, (B) 1, (C) -, - (D), 5. The vlues of k for which x 4x + k = 0 hs distinct rel roots re (A) k 4 (B) k > 4 (C) k < 4 (D) k 4 6. If -5 is root of the qudrtic eqution x + px 15 = 0 nd the qudrtic eqution p(x + x) + k = 0 hs equl roots then k = (A) 7 4 (B) - 7 4 (C) 4 7 (D) - 4 7 7. If the roots of the eqution (b c)x + (c )x + ( b) = 0 re equl then (A), b, c re in A.P. (B)., c, b re in A.P. (C) b,, c re in A.P. (D) none of these 8. A plne left 0 minutes lter thn the scheduled time nd in order to rech its destintion 1500 km wy in time, it hs to increse its speed by 50 km/hr from its usul speed. It s usul speed is (A) 750 km/hr (B) 1000 km/hr (C) 500 km/hr (D) 00 km/hr 9. If p nd q re roots of x + px + q = 0 then (A) p = 1, q = - (B) p = 0, q = 1 (C) p = -, q = 0 (D) p = -, q = 1 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 10 ASCENTPlus ASCENT Progrm & School 10. If, re roots of x + bx + c = 0 then (1 + ) (1 + ) is equl to (A) b c (B) b c (C) b c (D) b c 11. If, re roots of x 7x + 1 = 0 then the eqution whose roots re, is (A) (x) 7(x) + 1 = 0 x x (B) 7 1 0 (C) (x ) 7(x ) + 1 = 0 (D) (x + ) 7(x + ) + 1 = 0 1. If, re roots of x x 6 = 0 then eqution whose roots re +, + is (A) 4x + 49x + 118 = 0 (B) 4x 49x + 118 = 0 (C) 4x 49x 118 = 0 (D) x 49x + 118 = 0 1. Mtch the following: The condition tht one root of x + bx + c = 0 is A) reciprocl of the other root 1) nb = c(1 + n) B) nth power of the other root ) c( + c) + b = bc C) squre of the other root 1 1 n c n1 n c n1 b 0 ) D) n times the other root 4) c = 5) c( + c) = bc + b 14. Mtch the following: The roots of the eqution 1) x x + 4 = 0 re A) rel ) 6x x = 0 re B) rel nd equl ) x + x + = 0 re C) rel nd distinct 4) (b c) x + (c )x + ( b) = 0 re D) imginry Fill in the Blnks: 15. The vlues of m for which the eqution x mx + 7m 1 = 0 hs equl roots re. 16. The vlues of x for which x / + x 1/ = 0 re Short nswer questions: 17. Find the vlue of k for which one root is squre of the other root of x x k = 0. 18. If, re roots of x + bx + c = 0 then find the vlue of. 19. Write the condition tht the roots of x + bx + c = 0 re rel. 0. Find the nture of roots of the eqution 9x 0x + 5 = 0. FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

ASCENTPlus ASCENT Progrm & School Qudrtic Equtions- 10 ANSWERS TO ASSIGNMENT PROBLEMS SUBJECTIVE 1. x = b b or SECTION A. b b or. x = 0 or + b 4. x = - 1 or q p 5. m = 1 8. k = 16 9. 50 km/hr, 40 km /hr 10. 1, 4 or -1, -4 SECTION B 5. A = -, B = 77 9. k = - 1 FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.

Qudrtic Equtions- 104 ASCENTPlus ASCENT Progrm & School OBJECTIVE 1. C. B. B 4. A 5. C 6. A 7. A 8. A 9. A 10. B 11. B 1. B 1. A 4, B, C, D 1 14. 1 B, C, D, 4 A 15. m = or m = 4 16. x = 0 17. k = 5 18. c 19. b 4c 0 0. rel nd equl FIITJEE Ltd., ICES House, 9 A, Klu Sri, Srvpriy Vihr, New Delhi -110016, Ph 6515949, 656949, Fx:011-65194.