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[Audio] 157 Quadratic Equations and Inequations 157 Chapter 4 CONTENTS 4.1 Polynomial 4.2 Types of quadratic equation 4.3 Solution of quadratic equation 4.4 Nature of roots 4.5 Root under particular conditions Euclid 4.6 Relation between roots and coefficients 4.7 Biquadratic equation 4.8 Condition for common roots 4.9 Properties of quadratic equation 4.10 Quadratic expression 4.11 Wavy curve method 4.12 Position of roots of a quadratic equation 4.13 Descartes rule of signs 4.14 Rational algebraic inequations 4.15 Algebraic interpretation of rolle's Theorem The Babylonians knew of quadratic equations some 4000 years ago. The Greek mathematician Euclid (300 B.C.) gives several quadratic equation while solving geometrical problems,. Aryabhatta (476 A.D.) gives a rule to sum the geometric series which involves the solution of the quadratic equations Brahmagupta (598 A.D.) provides a rule for the solution of the quadratic equations which is very much the quadratic formula. Mahavira around 850 A.D. proposed a problem involving the use of quadratic equation and its solution. It was Sridhara, an Indian mathematician, around 900 A.D. who was the first to give an algebraic solution of the general equation ax2 + bx + c ac b b x = 0 a  0, showing the roots to be 2  4    a 2 4.16 Equations and inequations on containing absolute value Assignment (Basic and Advance Level) The first important treatment of a quadratic equation, by factoring, is found in Harriot's works in approximately 1631 A.D. Answer Sheet of Assignment.

[Audio] 158 158 Quadratic Equations and Inequations 4.1 Polynomial. Algebraic expression containing many terms of the form n cx , n being a non-negative n 2 3 1 integer is called a polynomial. i.e., n  n a x x a a x a x a x a f x         n 2 1 0 ...... ( ) , where x is a 3 1 variable, an a a a ....... , , 2 1 0 are constants and an  0 Example : 3 5 7 3 4 2 3 4     x x x x , 5 3 3 2 3    x x x . (1) Real polynomial : Let an a a a ....... , , 2 1 0 be real numbers and x is a real variable. 2 Then n anx a x a x a x a f x       ...... ) ( 3 2 1 0 is called real polynomial of real variable x with 3 real coefficients. Example : 1 2 ,4 5 4 3 2 2 3      x x x x x etc. are real polynomials. (2) Complex polynomial : If an a a a ....... , , 2 1 0 be complex numbers and x is a varying complex number. 2 Then n anx a x a x a x a f x       ...... ) ( 3 2 1 0 is called complex polynomial of complex.

[Audio] Quadratic Equations and Inequations 159 (1) Purely quadratic equation : A quadratic equation in which the term containing the first degree of the unknown quantity is absent is called a purely quadratic equation. i.e. 0 2  c  ax where a, c  C and a  0 (2) Adfected quadratic equation : A quadratic equation which contains terms of first as well as second degrees of the unknown quantity is called an adfected quadratic equation. i.e. 0 2    c bx ax where a, b, c  C and a  0, b  0. (3) Roots of a quadratic equation : The values of variable x which satisfy the quadratic equation is called roots of quadratic equation. Important Tips  An equation of degree n has n roots, real or imaginary.  Surd and imaginary roots always occur in pairs in a polynomial equation with real coefficients i.e. if 2 – 3i is a root of an equation, then 2 + 3i is also its root. Similarly if 3 2  is a root of given equation, then 3 2  is also its root.  An odd degree equation has at least one real root whose sign is opposite to that of its last term (constant term), provided that the coefficient of highest degree term is positive.  Every equation of an even degree whose constant term is negative and the coefficient of highest degree term is positive has at least two real roots, one positive and one negative. 4.3 Solution of Quadratic Equation. (1) Factorization method : Let 0 ) )( ( 2         x a x c bx ax . Then x  and x   will satisfy the given equation. Hence, factorize the equation and equating each factor to zero gives roots of the equation. Example : 0 1 2 3 2   x  x  0 )1 1)(3 (    x x x  ,1 1 / 3 (2) Hindu method (Sri Dharacharya method) : By completing the perfect square as 0 2    c bx ax  0 2    a c a x b x 2 2 2  ac b 4  Adding and subtracting 2  4    ac b b x 2 a b , 0 2  a b x which gives, a 2 2      a           4       Hence the quadratic equation 0 2    c bx ax (a  0) has two roots, given by ac b b ac b b 2  4     2  4     , a a 2 2 Note :  Every quadratic equation has two and only two roots. 4.4 Nature of Roots..

[Audio] 160 Quadratic Equations and Inequations In quadratic equation 0 2    c bx ax , the term ac b 2  4 is called discriminant of the equation, which plays an important role in finding the nature of the roots. It is denoted by  or D. (1) If a, b, c  R and a  0, then : (i) If D < 0, then equation 0 2    c bx ax has non-real complex roots. D b (ii) If D > 0, then equation 0 2    c bx ax has real and distinct roots, namely a 2     , D b a 2     and then ) )( ( 2        x a x c bx ax …..(i) (iii) If D = 0, then equation 0 2    c bx ax has real and equal roots a b 2      and then 2 2 ) (     a x c bx ax …..(ii) To represent the quadratic expression c bx ax  2  in form (i) and (ii), transform it into linear factors. (iv) If D  0, then equation 0 2    c bx ax has real roots. (2) If a, b, c  Q, a  0, then : (i) If D > 0 and D is a perfect square  roots are unequal and rational. (ii) If D > 0 and D is not a perfect square  roots are irrational and unequal. (3) Conjugate roots : The irrational and complex roots of a quadratic equation always occur in pairs. Therefore (i) If one root be    i then other root will be    i . (ii) If one root be    then other root will be    . (4) If D1 and D2 be the discriminants of two quadratic equations,then (i) If 0 2 1 D  D  , then (a) At least one of 1 D and D2  0 . (b) If D1  0 then D2  0 (ii) If 0 2 1 D  D  , then (a) At least one of 1 D and D2  0 . (b) If D1  0 then D2  0 . 4.5 Roots Under Particular Conditions. For the quadratic equation 0 2    c bx ax . (1) If b  0  roots are of equal magnitude but of opposite sign. (2) If c  0  one root is zero, other is  b / a . (3) If b  c  0  both roots are zero. (4) If a  c  roots are reciprocal to each other..

[Audio] Quadratic Equations and Inequations 161   0 0 (5) If c a  roots are of opposite signs.   c a 0 0       0 0 0 (6) If c b a  both roots are negative, provided D  0 .    c b a 0 0 0       0 0 0 (7) If c b a  both roots are positive, provided D  0 .    c b a 0 0 0    (8) If sign of a = sign of b  sign of c  greater root in magnitude, is negative. (9) If sign of b = sign of c  sign of a  greater root in magnitude, is positive. (10) If  0   c b a  one root is 1 and second root is c/a. (11) If  0   c b a , then equation will become an identity and will be satisfied by every value of x. (12) If a  1 and b, c  I and the root of equation 0 2    c bx ax are rational numbers, then these roots must be integers. Important Tips  If an equation has only one change of sign, it has one +ve root and no more.  If all the terms of an equation are +ve and the equation involves no odd power of x, then all its roots are complex. Example: 1 Both the roots of given equation 0 ) )( ( ) )( ( ) )( (          a c x x c b x x b a x x are always [MNR 1986; IIT 1980; Kurukshetra CEE 1998] (a) Positive (b) Negative (c) Real (d) Imaginary Solution: (c) Given equation 0 ) )( ( ) )( ( ) )( (          a c x x c b x x b a x x can be re-written as 0 ) ( ) 2( 3 2        ca bc ab c x b a x ] 4[ )] 3( ) 4[( 2 2 2 2 ac bc ab c b a ca bc ab c b a D             0 ) ] ( ) ( ) 2[( 2 2 2        a c c b b a Hence both roots are always real. Example: 2 If the roots of 0 ) ( ) ( ) ( 2       b a a x c c x b are equal then a  c  [Kurukshetra CEE 1992] (a) 2b (b) 2 b (c) 3b (d) b Solution: (a)  0      b a a c c b Hence one root is 1. Also as roots are equal, other root will also be equal to 1. b a b a    1 1.  c b b a     c a b   2 Also c b   .   c b are equal in magnitude but opposite in sign, then  )  ( q p 1 1 1     Example: 3 If the roots of equation r q x p x [Rajasthan PET 1999] (a) 2r (b) r (c) – 2r (d) None of these Solution: (a) Given equation can be written as 0 ) ] ( [ 2 ) ( 2        q r p pq r x q p x Since the roots are equal and of opposite sign,  Sum of roots = 0  0 2 ) (     r q p  r q p  2  Example: 4 If 3 is a root of 0 24 2  x  kx  , it is also a root of [EAMCET 2002] (a) 0 5 2    k x x (b) 0 5 2  {  k x x.

[Audio] 162 Quadratic Equations and Inequations Solution: (c) Equation 0 24 2  x  kx  has one root as 3,  0 24 3 32   k   k  5 Put x  3 and k  5 in option Only (c) gives the correct answer i.e.  0 9 15 32     0  0 Example: 5 For what values of k will the equation 0 2 ) 7(3 3 ) 2 1( 2      k k x x have equal roots [MP PET 1997] (a) 1, –10/9 (b) 2, –10/9 (c) 3, –10/9 (d) 4, –10/9 Solution: (b) Since roots are equal then 2 ) (3 7.1.4 3 )] [ 2 1( 2 k k      k k k 14 21 6 9 1 2      0 20 8 9 2  k  k  Solving, we get k  ,2 10 / 9 4.6 Relations between Roots and Coefficients. (1) Relation between roots and coefficients of quadratic equation : If  and  are the roots of quadratic equation 0 2    c bx ax , (a  0) then t of coefficien a b S          x x Sum of roots t of 2 coefficien constant term . x a c P       Product of roots t of 2 coefficien If roots of quadratic equation 0 2    c bx ax (a  0) are  and  then 2 D 2      ac b          4 4 ) ( ) ( (i) a a 2 ac b            2 2 2 2 2 ) ( (ii) 2 a 2 D b ac b b                  2 2 2 4 4 ) ) ( ( (iii) 2 2 a a 2 ac b b                3 3 3 ) 3 ( ) ( 3 ) ( (iv) 3 a 2 2 ac b ac b                            2 2 3 3 3 4 ) ( } ) 2 ) {( (vi) 2 2 a a c   ac b    2 2 ac b ac b b               2 2 2 2 4 4 4 ) 2 ( ) )( ( (vii) 4 a 2 ac b              2 2 2 ) ( (viii) 2 a        ac b 2 2 ) ( 2 2 2 2             (ix) ac 2 2 ) ( (x) 2 a   bc           2 2 2 2 2 2 2 2        4 4 2 2 2 2 ) ( (xi) 2 2 2 2 2 2         c a   a c b D a                .

[Audio] Quadratic Equations and Inequations 163 (2) Formation of an equation with given roots : A quadratic equation whose roots are  and  is given by 0 ) )( (      x x  0 ) ( 2        x x i.e. 0 (product of roots) (sum of roots) 2    x x  0 2    P Sx x (3) Equation in terms of the roots of another equation : If ,  are roots of the equation 0 2    c bx ax , then the equation whose roots are (i) –, –  0 2    c bx ax (Replace x by – x) (ii)   1 / ,1 /  0 2    a bx cx (Replace x by 1/x) (iii) n  n  , ; n  N  0 ) ( ) ( 1/ 2 1/    c b x a x n n (Replace x by n x 1 / ) (iv) k, k  0 2 2    k c kbx ax (Replace x by x/k) (v) k  , k    0 ) ( ) ( 2      c k b x k a x (Replace x by (x – k))  ,   0 2 2    c kbx k ax (Replace x by kx) (vi) k k (vii) n n 1 / 1 / ,   ; n  N  0 ) ( ) ( 2    c b x a x n n (Replace x by n x ) (4) Symmetric expressions : The symmetric expressions of the roots ,  of an equation are those expressions in  and , which do not change by interchanging  and . To find the value of such an expression, we generally express that in terms of    and . Some examples of symmetric expressions are : (i) 2 2    (ii) 2 2      (iii)   1  1 (iv)      2 2   (v)     2 2  (vi)   (vii) 3 3    (viii) 4 4                 4.7 Biquadratic Equation. If , , ,  are roots of the biquadratic equation 0 2 3 4      e dx cx bx ax , then 2 )1 ( . . .          b a S / 1           , a c a c S          2 3             or c a S / ) )( ( 2             , d a a d S / )1 ( 3 4 )1 ( . . .     or d a S / ) ( ) ( 3             and a e a e S     4 Example: 6 If the difference between the corresponding roots of 0 2    b ax x and 0 2    a bx x is same and a  b , then [AIEEE 2002] (a) ab  4  0 (b) a b 4  0 (c) a b 4  0 (d) a b  4  0 Solution: (a)     a ,   b  b a 2  4     and     b ,   a  a b 2  4    According to question,         a b b a 4 4 2 2     ab  4  0 Example: 7 If the sum of the.

[Audio] 164 Quadratic Equations and Inequations Solution: (c) As given, if ,  be the roots of the quadratic equation, then 2 2    c a a b ac b 2 / / 2 2 2 2 ) ( 1 1 2 2          2 2 2  2 2 a b          a c c / 2 2 bc ab 2 2 2  2 b a   2 c b c a      2 2 2 2 bc ab a c    a c a b c b ac c a , , are in H.P. b a a b c b a c , , are in A.P.  b c Example: 8 Let ,  be the roots of 0 2    p x x and ,  be root of 0 4 2    q x x . If , , ,  are in G.P., then the integral value of p and q respectively are [IIT Screening 2001] (a) – 2, – 32 (b) – 2, 3 (c) – 6, 3 (d) – 6, – 32 Solution: (a)     1 ,   p ,     4 ,   q Since , , ,  are in G.P.       / / /   r   r  1    1 ) 1(  r   , 4 ) ( 3 2   r  r  4 ) . 2 1(  r   r So r2  4  r  2 If r  2 , 1 2        1 / 3 and r  2 , 1 2        1 But I p     1 ,2      r  p  2 , 32 2) (1 5 2 5      r q  Example: 9 If 1 – i is a root of the equation 0 2    b ax x , then the values of a and b are [Tamil Nadu Engg. 2002] (a) 2, 1 (b) – 2, 2 (c) 2, 2 (d) 2, – 2 Solution: (b) Since 1i is a root of 0 2    b ax x .  1i is also a root. Sum of roots  a i i      1 1  a  2 Product of roots  b i i    ) )(1 1(  b  2 Hence a  2 , b  2 Example: 10 If the roots of the equation 0 16 5 2  x  x  are ,  and the roots of equation 0 2    q px x are 2 2    ,  / 2 , then [MP PET 2001] (a) 56 ,1    q p (b) 56 ,1     q p (c) 56 ,1   q p (d) 56 ,1    q p Solution: (b) Since roots of the equation 0 16 5 2  x  x  are ,  .  16 ,5       and  p   2 2 2      p    2 2 ) ( 2       p   2 16 2(16) 25  p  1   2 ) ( 2 2      q   ] 2 2 ) [( 2       q  32 8) (25  q  56 and   q      is  and  Example: 11 If    , but 3 5 ,3 5 2 2         , then the equation whose roots are  [EAMCET 1989; AIEEE 2002] (a) 0 3 5 2    x x (b) 0 3 19 3 2    x x (c) 0 3 12 3 2    x x (d) None of these 2     3 5       2 Solution: (b)    3 5 3 5 2 2        S         3 5  {    .