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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  {    .

[Audio] Quadratic Equations and Inequations 165 ∵ 0 1 3 19 2    x x  0 3 19 3 2    x x Example: 12 Let ,  be the roots of the equation c b a x x    ) )( ( , c  0 , then the roots of the equation 0 ) )( (     c x x   are [IIT 1992; DCE 1998, 2000; Roorkee 2000] (a) a, c (b) b, c (c) a, b (d) a, d Solution: (c) Since ,  are the roots of c b a x x    ) )( ( i.e. of 0 ) ( 2      c ab b x a x      a  b  a  b     and   ab  c  c ab     a, b are the roots of 0 ) ( 2      c x x     0 ) )( (     c x x   Hence (c) is the correct answer Example: 13 If  and  are roots of the equation 0 2    b ax x and n n Vn     , then [Rajasthan PET 1995; Karnataka CET 200 (a) 1 1     n n n b V aV V (b) 1 1     n n n a V b V V (c) 1 1     n n n b V a V V (d) 1 1     n n n a V b V V Solution: (a) Since  and  are roots of equation, 0 2    b ax x , therefore     a ,   b     Now, ) ( ) )( ( 1 1 1 1         n n n n n n 1 Vn           1 1 . .     n n n b V a V V Example: 14 If one root of the equation 0 2    q px x is the square of the other, then [IIT Screening 2004] (a) 0 )1 (3 2 3     p q q p (b) 0 3 ) 1( 2 3     p q q p (c) 0 )1 (3 2 3     p q q p (d) 0 3 ) 1( 2 3     p q q p Solution: (d) Let  and 2  be the roots then  p  2  , .2  q Now ) ( 3 ) ( 2 3 6 3 2 3              pq q q p 3 2 3      0 3 ) 1( 2 3     p q q p Example: 15 Let  and  be the roots of the equation 0 1 2  x  x  , the equation whose roots are 19, 7 is[IIT 1994; Pb. CET (a) 0 1 2  x  x  (b) 0 1 2  x  x  (c) 0 1 2  x  x  (d) 0 1 2  x  x  4 1 1           i x Solution: (d) Roots of 0 1 2  x  x  are 2 , 2 3 1 , 2 Take 2 ,        2 14 2 7 7 19 19 ) ( , w w w w w         Required equation is 0 1 2  x  x  1 Example: 16 If one root of a quadratic equation is , then the equation is [Rajasthan PET 1987]  5 2 (a) 0 1 4 2    x x (b) 0 1 4 2    x x (c) 0 1 4 2    x x (d) None of these 5 2  ,.

[Audio] 166 Quadratic Equations and Inequations 2  The condition for only one root common is ) )( ( ) ( 1 2 2 1 1 2 2 1 1 2 1 2 a b a b b c b c c a c a     1 1 1 (2) Both roots are common: Then required condition is a a   . b b c c 2 2 2 Important Tips  To find the common root of two equations, make the coefficient of second degree term in the two equations equal and subtract. The value of x obtained is the required common root.  Two different quadratic equations with rational coefficient can not have single common root which is complex or irrational as imaginary and surd roots always occur in pair. Example: 17 If one of the roots of the equation 0 2    b ax x and 0 2    a bx x is coincident. Then the numerical value of ) ( a  b is [IIT 1986; Rajasthan PET 1992; EAMCET 2002] (a) 0 (b) – 1 (c) (d) 5 Solution: (b) If  is the coincident root, then 0 2    b a  and 0 2    a b  1 2   2 2   a b a b b a     ) ( 2   a  b  ,   1  1 ) (    b a  1 ) ( a  b   Example: 18 If a, b, c are in G.P. then the equations 0 2 2    c bx ax and 0 2 2    f ex dx have a common root if a d , , are in b e c f [IIT 1985; Pb. CET 2000; DCE 2000] (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (a) As given, ac b 2   0 2 2    c bx ax can be written as 0 2 2    c acx ax  0 ) ( 2   c ax  a c x   This must be common root by hypothesis  f a c e a c d So it must satisfy the equation, 0 2 2    f ex dx  0 2         e 2 2     b e a d  2  c f a d c f c e a c a c . a d , , are in A.P. b e Hence c f 4.9 Properties of Quadratic Equation. (1) If f(a) and f(b) are of opposite signs then at least one or in general odd number of roots of the equation 0 ( ) f x  lie between a and b. Y Y f(a) = +ve f(a) = +ve x = b A x = b x = a O X x = a B X f(b) = – ve f(b) = – ve (2) If ( ) ( ) f b f a  then there exists a point c between a and b such that 0 ( ) f c  , b c a   . Y Q P X O A a B b.

[Audio] Quadratic Equations and Inequations 167 As is clear from the figure, in either case there is a point P or Q at x  c where tangent is parallel to x-axis i.e. 0 ( ) f x  at x  c . (3) If  is a root of the equation 0 ( ) f x  then the polynomial f(x) is exactly divisible by ) ( x  or ) ( x  is factor of f(x) . 2 2    c b x a x are in the same (4) If the roots of the quadratic equations 0 2    c bx ax , 0 2 2  2   2 1 / / a c a c b b  . 2 2  i e  then 2 2 1 1 1 . .   1 2   ratio   (5) If one root is k times the other root of the quadratic equation 0 2    c bx ax then k 2 1 2) (   . k ac b Example: 19 The value of 'a' for which one root of the quadratic equation 0 2 1) (3 3) 5 ( 2 2       x a x a a is twice as large as the other is [AIEEE 2003] (a) 2/3 (b) – 2/3 (c) 1/3 (d) – 1/3 Solution: (a) Let the roots are  and 2 2   Now, a   ,    a  ,    a a a a a a a a 3 5 3 1 2 2      3 5 2 . 2 2   3 5 3 1 3 2     3 5 2 2 2 2 2   3 ) 1( 3 ) 1(    a  9 a  a a a a 6 9 1 27 45 9 2 2       26 39 a   a  2 / 3 2      9 1 2 2 2 2    a a a a 3) 5 ( 3 5 2 a a 3 5     4.10 Quadratic Expression. An expression of the form c bx ax  2  , where a, b, c  R and a  0 is called a quadratic expression in x. So in general, quadratic expression is represented by c bx ax f x    2 ( ) or c bx ax y    2 . (1) Graph of a quadratic expression : We have ) ( 2 f x c bx ax y     2 2   a b x a y  2 4          a b a x a D y          2 4 2 a D       Now, let Y a D y   4 and a b x X  2  Y  a.X 2  a Y X 2  1 (i) The graph of the curve y  f(x) is parabolic. (ii) The axis of parabola is X  0 or 0 2   a b x i.e. (parallel to y-axis). (iii) (a) If a > 0, then the parabola opens upward..

[Audio] 168 Quadratic Equations and Inequations (b) If a 0, D < 0 a < 0, D < 0 x axis (iv) Intersection with axis (a) x-axis: For x axis, y  0  0 2    c bx ax  a D b x 2    For D > 0, parabola cuts x-axis in two real and distinct points i.e. a D b x 2    . For D = 0, parabola touches x-axis in one point, a b x   / 2 . a < 0 D > 0 xaxis xaxis a > 0 D > 0 a < 0 D = 0 xaxis xaxis a > 0 D = 0 For D < 0, parabola does not cut x-axis(i.e. imaginary value of x). a < 0 D 0 D < 0 (b) y-axis : For y axis x  0 , y  c (2) Maximum and minimum values of quadratic expression : Maximum and minimum value of quadratic expression can be found out by two methods : (i) Discriminant method : In a quadratic expression c bx ax  2  ..

[Audio] Quadratic Equations and Inequations 169 (a) If a > 0, quadratic expression has least value at a b x   / 2 . This least value is given by b ac 4 2    . a a D 4 4 (b) If a < 0, quadratic expression has greatest value at a b x   / 2 . This greatest value is b ac 4 2    . given by a D a 4 4 (ii) Graphical method : Vertex of the parabola Y  aX2 is X  0 , Y  0 i.e., 0 2   a b x , 0 4   a D y  a b x   / 2 , a D y / 4   Hence, vertex of c bx ax y    2 is / 4 ) / 2 , ( a D a b   (a) For a > 0, f(x) has least value at a b x   2 . This least value is given a > 0 by a D a b f 4 2         . (b) For a < 0, f(x) has greatest value at a b x   / 2 . This greatest value is verte x verte x given by a D a b f 4 2         . a < 0 (3) Sign of quadratic expression : Let c bx ax f x    2 ( ) or c bx ax y    2 Where a, b, c  R and a  0, for some values of x, f(x) may be positive, negative or zero. This gives the following cases : (i) a > 0 and D < 0, so 0 ( ) f x  for all x  R i.e., f(x) is positive for all real values of x. (ii) a < 0 and D < 0, so 0 ( ) f x  for all x  R i.e., f(x) is negative for all real values of x. (iii) a > 0 and D = 0 so, 0 ( ) f x  for all x  R i.e., f(x) is positive for all real values of x except at vertex, where 0 ( ) f x  . (iv) a < 0 and D = 0 so, 0 ( ) f x  for all x  R i.e. f(x) is negative for all real values of x except at vertex, where 0 ( ) f x  . (v) a > 0 and D > 0 Let 0 ( ) f x  have two real roots  and  ) (    , then 0 ( ) f x  for all ) ( , , ) (       x and 0 ( ) f x  for all ) ( , x    . (vi) a  0 and D  0 Let 0 ( ) f x  have two real roots  and  ) (    , Then 0 ( ) f x  for all ) ( , , ) (       x and 0 ( ) f x  for all ) ( , x    Example: 20 If x be real, then the minimum value of 17 8 2 x  x  is [MNR 1980] (a) – 1 (b) 0 (c) 1 (d) 2 4 2 64 1( )(17) 4 b ac       a Solution: (c) Since a 1  0 therefore its minimum value is 1 4 4 4 4 2   Example: 21 If x is real, then greatest and least values of x x are [IIT 1968; Rajasthan PET 1988] 2   x x 1 1 (a) 3, –1/2 (b) 3, 1/3 (c) – 3, –1/3 (d) None.

[Audio] 170 Quadratic Equations and Inequations 2    Solution: (b) Let x x y 2   x x 1 1 0 )1 ( )1 ( )1 2(       y x y y x ∵ x is real, therefore 0 4 2 b  ac    y y  3 3 1  y      0 )1 1)( 4( )1 ( 2      y y y  0 3 10 3 2    y y  0 3) 1)( (3    y y  0 3) 3 ( 1     Thus greatest and least values of expression are 3, 1/3 respectively. Example: 22 If f(x) is quadratic expression which is positive for all real value of x and ( ) ( ) ( ) ( ) x f f x f x g x      . Then for any real value of x [IIT 1990] (a) 0 ( ) g x  (b) 0 ( ) g x  (c) 0 ( ) g x  (d) 0 ( ) g x  Solution: (b) Let c bx ax f x    2 ( ) , then 2 ) ( 2 ) ( 2 2 ( ) 2 2 a c b a x b ax a b ax c bx ax g x             ∵ 0 ( ) f x  . Therefore 0 4 2 b  ac  and a  0 Now for g(x), Discriminant 0 4 ) 4 ( 8 4 4 4 4 2 ) 4 ( 2 ) ( 2 2 2 2 2 2                a ac b a ac ab ab a b a c a b a b as 0 4 2 b  ac  Therefore sign of g(x) and a are same i.e. 0 ( ) g x  . Example: 23 If ,  ) (    are roots of the equation 0 2    c bx x where ) 0 ( b c   then [IIT Screening 2000] (a) 0     (b) | | 0       (c)     0 (d)       | 0 | Solution: (b) Since 0 0 0 (0)      c c f  Roots will be of opposite sign, ve b        (b > 0) It is given that    So,     ve is possible only when   |  |         | |,0 ,0  | | 0       4.11 Wavy Curve Method. k Let n n k k k k a x a x a x a x a x f x ) ( ) ) .......( ) ( ) ( ( ) ( 1 3 2 1  …..(i) n n 1 3 2 1        Where N k k k k ..., n  , , 3 2 1 and an a a a ,......, , , 3 2 1 are fixed natural numbers satisfying the condition n n a a a a a     1 3 2 1 ..... First we mark the numbers an a a a ,......, , , 3 2 1 on the real axis and the plus sign in the interval of the right of the largest of these numbers, i.e. on the right of n a . If n k is even then we put plus sign on the left of n a and if n k is odd then we put minus sign on the left of n a . In the next interval we put a sign according to the following rule : When passing through the.

[Audio] Quadratic Equations and Inequations 171 Let c bx ax f x    2 ( ) , where a, b, c  R be a quadratic expression and , 1, 2 k k k be real numbers such that 2 1 k k  . Let ,  be the roots of the equation 0 ( ) f x  i.e. 0 2    c bx ax . Then D b D b a 2     , a 2     where D is the discriminant of the equation. (1) Condition for a number k (If both the roots of f(x) = 0 are less than k) a > 0 (–b/2a, – D/4a) f(k ) k   xaxis – b/2a   k – b/2a xaxis a < 0 f(k ) (–b/2a, – D/4a) (i) D  0 (roots may be equal) (ii) 0 ( ) a f k  (iii) a b k   / 2 , where    (2) Condition for a number k (If both the roots of f(x) = 0 are greater than k) a > 0 (–b/2a, – D/4a) f(k ) k   – b/2a   k xaxis – b/2a xaxis a < 0 f(k ) (–b/2a, – D/4a) (i) D  0 (roots may be equal) (ii) 0 ( ) a f k  (iii) a b k   / 2 , where    a > 0 (–b/2a, – D/4a) (3) Condition for a number k (If k lies between the roots of f(x) = 0) k     k xaxis xaxis a < 0 (–b/2a, – D/4a) (i) D  0 (ii) 0 ( ) a f k  , where    (4) Condition for numbers k1 and k2 (If exactly one root of f(x) = 0 lies in the interval (k1, k2)) a > 0 k1 f(k   f(k1 ) 2) (k2 x axis )   k1 f(k f(k1 ) 2) (k2 ) xaxis a < 0.

[Audio] 172 Quadratic Equations and Inequations (i) D  0 (ii) 0 ) ) ( ( 2 1  f k f k , where    . (–b/2a, – D/4a) a > 0 f(k k1 k2 f(k1 ) 2)     k1 k2 a < 0 (–b/2a, – D/4a) (5) Condition for numbers k1 and k2 (If both roots of f(x) = 0 are confined between k1 and k2) (i) D  0 (roots may be equal) (ii) 0 ) ( 1  a f k (iii) 0 ) ( a f k2  (iv) 2 1 / 2 k a b k    , where    and 2 1 k k  a > 0   k1 k2 x-axis k1 k2 a < 0   f(k xaxis f(k1 ) 2) (6) Condition for numbers k1 and k2 (If k1 and k2 lie between the roots of f(x) = 0) (i) D  0 (ii) 0 ) ( 1  a f k (iii) 0 ) ( a f k2  , where    Example: 24 If the roots of the equation 0 3 2 2 2      a a ax x are real and less than 3, then [IIT 1999; MP PET 2000] (a) a < 2 (b) 3 2  a  (c) 4 3  a  (d) a  4 Solution: (a) Given equation is 0 3 2 2 2      a a ax x If roots are real, then D  0  0 3) 4( 4 2 2     a a a  a 3  0  a 3  0  a  3 As roots are less than 3, hence 0 (3)  f 0 3 6 9 2      a a a  0 6 5 2  a  a   0 3) 2)( (    a a  3 ,2   a a . Hence a < 2 satisfy all the conditions. Example: 25 The value of a for which 0 )1 ( )1 2(2 2 2      a a x a x may have one root less than a and another root greater than a are given by [UPSEAT 2001] (a) 0 1  a  (b) 0 1    a (c) a  0 (d) a  0 or a  1 Solution: (d) The given condition suggest that a lies between the roots. Let )1 ( )1 2(2 2 ( ) 2      a a x a x f x For 'a' to lie between the roots we must have Discriminant  0 and 0 ( ) f a  Now, Discriminant  0 0 )1 8 ( )1 4(2 2     a a a  0 1 / 2) 8( 2  a  a  which is always true. Also 0 ( ) f a   0 )1 ( )1 2 (2 2 2      a a a a a  0 2    a a  0 2 a  a   0 ) 1(  a  a  a  0 or a  1.

[Audio] Quadratic Equations and Inequations 173 4.13 Descarte's Rule of Signs. The maximum number of positive real roots of a polynomial equation 0 ( ) f x  is the number of changes of sign from positive to negative and negative to positive in f(x). The maximum number of negative real roots of a polynomial equation 0 ( ) f x  is the number of changes of sign from positive to negative and negative to positive in ( x) f  . Example: 26 The maximum possible number of real roots of equation 0 5 4 6 2 5     x x x is [EAMCET 2002] (a) 0 (b) 3 (c) 4 (d) 5 Solution: (b) 0 5 4 6 ( ) 2 5      x x x f x + – – + 2 changes of sign  maximum two positive roots. 5 4 6 ) ( 2 5       x x x x f – – + + 1 changes of sign  maximum one negative roots.  total maximum possible number of real roots = 2 + 1 = 3. 4.14 Rational Algebraic Inequations. (1) Values of rational expression P(x)/Q(x) for real values of x, where P(x) and Q(x) are quadratic 2 c b x x a 1 1 1   for expressions : To find the values attained by rational expression of the form 2   c b x x a 2 2 2 real values of x, the following algorithm will explain the procedure : Algorithm Step I: Equate the given rational expression to y. Step II: Obtain a quadratic equation in x by simplifying the expression in step I. Step III: Obtain the discriminant of the quadratic equation in Step II. Step IV: Put Discriminant  0 and solve the inequation for y. The values of y so obtained determines the set of values attained by the given rational expression. (2) Solution of rational algebraic inequation: If P(x) and Q(x) are polynomial in x, then the x P x P P x are known as rational algebraic inequations. x Q ( )    x Q x Q ( ) x  Q inequation 0 ) ( P x and 0 ) ( ( ) ,0 ) ( ( ) ,0 ) ( To solve these inequations we use the sign method as explained in the following algorithm. Algorithm Step I: Obtain P(x) and Q(x). Step II: Factorize P(x) and Q(x) into linear factors. Step III: Make the coefficient of x positive in all factors. Step IV: Obtain critical points by equating all factors to zero. Step V: Plot the critical points on the number line. If there are n critical points, they divide the number line into (n + 1) regions..

[Audio] 174 Quadratic Equations and Inequations P x bears positive sign and in other ( ) x Q Step VI: In the right most region the expression ) ( regions the expression bears positive and negative signs depending on the exponents of the factors. 4.15 Algebraic Interpretation of Rolle's Theorem. Let f(x) be a polynomial having  and  as its roots, such that    . Then, 0 ( ) ( )     f f . Also a polynomial function is everywhere continuous and differentiable. Thus f(x) satisfies all the three conditions of Rolle's theorem. Consequently there exists ) ( ,     such that 0 ( ) f   i.e. 0 ( ) f x  at x   . In other words x   is a root of 0 ( ) f x  . Thus algebraically Rolle's theorem can be interpreted as follows. Between any two roots of polynomial f(x), there is always a root of its derivative f(x) . Lagrange's theorem : Let f(x) be a function defined on [a b] such that (i) f(x) is continuous on [a b] and f a f b f c     ( ) ( ) ( ) (ii) f(x) is differentiable on (a, b), then c  (a, b), such that a b Lagrange's identity : If R b b b a a a 3  2 1 3 2 1 , , , , , then : 2 2 2 2 2 1 ) ( ) ( ) ( ) ( ) )( ( a b a b a b a b a b a b a b a b a b b b b a a a              2 2 2 3 2 1 2 2 2 3 3 3 2 2 1 1 1 2 2 1 2 3 3 2 3 1 1 3 2 1 x , then [IIT 1987] 2   Example: 27 If 1   x x x 2 5 2 (a) 1 2     x (b) 1 2     x (c) 1 2     x (d) 1 2     x       2 5 2 2 2 2 2 2 1 x     x x x 2    2 3     x  0 1) 2)( 1)( 2 ( x x x Solution: (c) Given 0 1 x x x x  0 )1 2)( 1)( 2 (   x x x 2 5 2    x 2 / 3) 3(     2 / 3) (     x x x x x x  0 )1 2)(2 1)( ( x  0 )1 2)(2 1)( ( Equating each factor equal to 0, + + + – – –2/3 –2 –1 –1/2 We get 1/ 2 2 / ,3 ,1 ,2    x    1 / 2[ 2 / ,3 ] [1 ,2 ]     x    1 / 2 2 / 3     x or 1 2     x Example: 28 If for real values of x, 0 2 3 2    x x and 0 4 3 2    x x , then [IIT 1983] (a) 1 1    x (b) 4 1    x (c) 1 1    x or 4 2  x  (d) 4 2  x  Solution: (c) 0 2 3 2    x x or 0 2) 1)( (    x x + +  ) ( ,2 )1, (   x   …..(i) – 2 1 Again 0 4 3 2    x x or 0 1) 4)( (    x x + + –  x [ ,1 4] …..(ii) 4 –1 From eq. (i) and (ii), (24] ,1 )1 [  x    1 1    x or 4 2  x  Example: 29 If b  a , then the equation 0 1 ) )( (     b a x x , has [IIT Screening 2000] (a) Both roots in [a b] (b) Both roots in (– , a).

[Audio] Quadratic Equations and Inequations 175 (c) Both roots in (b, ) (d) One root in (– , a) and other in (b, +) Solution: (d) We have, 0 1 ) )( (     b a x x 0 1 ) )( (     b a x x  0 ) )( (    b a x x [∵ b > a] + + – b a [ ] , , [ ]      b a x , i.e. , ) (  a and (b, ).  1 Example: 30 The number of integral solution of x is [Orissa JEE 2002] 2  4 1  x 2 (a) 1 (b) 2 (c) 5 (d) None of these 2    2 4 1  x x  0 6)) (2 6))( (2 (      x x x  0 2 2    Solution: (c) 0 4 1 x  2 x 2 + +  6 2 6 2     x 2–6 – 2+6 Approximately, 4.4 4.0    x Hence, integral values of x are 0, 1, 2, 3, 4 Hence, number of integral solution = 5 Example: 31 If 0 6 3 2    c b a then at least one root of the equation 0 2    c bx ax lies in the interval [Kurukshetra CEE 2002; AIEEE 2002] (a) (0, 1) (b) (1, 2) (c) (2, 3) (d) (3, 4) Solution: (a) Let c bx ax f x     2 ( ) 2 3  cx bx ax f x dx f x      2 3 ( ) ) ( 6 3 2 Clearly 0 (0)  f , 0 6 0 2 3 1( )         c b a c b a f 6 Since, 0 1( ) (0)   f f . Hence, there exists at least one point c in between 0 and 1, such that 0 ( ) f  x  , by Rolle's theorem. Trick: Put the value of 0 ,2 ,3     c b a in given equation 0 2 3 2    x x  0 2 3 2   x x  0 2) (3  x  x 2 / 3 ,0   x x , which lie in the interval (0, 1) 4.16 Equation and Inequation containing Absolute Value. (1) Equations containing absolute values By definition, x x x   0 , if | | x x   0 if ,    Important forms containing absolute value : Form I: The equation of the form ( )| | ( )| ( )| | ( ) | g x f x g x f x    is equivalent of the system 0 ( ). ( ) f x g x  . Form II: The equation of the form ( ) |) ( ......| |) ( | |) ( | |) ( | 3 2 1 g x f x x f x f x f n     …..(i) Where ( ), ( ) ( )...... ( ), ), ( 3 2 1 g x f x x f x f x f n are functions of x and g(x) may be a constant..

[Audio] 176 Quadratic Equations and Inequations Equations of this form can be solved by the method of interval. We first find all critical points of ( ) ( )..... ), ( 2 1 f x x f x f n . If coefficient of x is +ve, then graph starts with +ve sign and if it is negative, then graph starts with negative sign. Then using the definition of the absolute value, we pass form equation (i) to a collection of system which do not contain the absolute value symbols. (2) Inequations containing absolute value By definition, |x| < a  a x a    (a > 0), x  a | |  a x a    , |x| > a  a x   or x  a and a x |  |  a x   or x  a Example: 32 The roots of 0 6 |2 | |2 | 2      x x are [UPSEAT 2003] (a) 0, 4 (b) –1, 3 (c) 4, 2 (d) 5, 1 Solution: (a) We have 0 6 |2 | |2 | 2      x x Let X x  |2  | 0 6 2  X  X  24 1 1     X    X  2 and X  3  3 ,2 2  2 |2 |  x  and 3 |2 |   x  , which is not possible.  x 2  2 or 2 2   x   x  4 or x  0 Example: 33 The set of all real numbers x for which 0 |2 2 |     x x x , is [IIT Screening 2002] (a) ) ( ,2 2) , (     (b) ) ( 2, 2) , (     (c) ) ,1( )1 , (     (d) ) ( 2,  Solution: (b) Case I: If x  2  0 i.e. x  2 , we get 0 2 2     x x x  0 2 2  x   0 2) 2)( (    x x + +  ( 2, ) 2) , (   x    –2 2 But x  2  ) ( 2 2) ,2 [   x    …..(i) Case II: x  2  0 i.e. x  2 , then 0 2 2     x x x  0 2 2 2    x x  0 1 )1 ( 2   x  . Which is true for all x  2) , ( x    …..(ii) From (i) and (ii), we get, ( 2, ) 2) , (   x    Example: 34 Product of real roots of the equation 0 9 | | 2 2  t x  x  (t  0) [AIEEE 2002] (a) Is always +ve (b) Is always –ve (c) Does not exist (d) None of these Solution: (c) Expression is always +ve, so 0 9 | | 2 2  t x  x  . Hence roots of given equation does not exist. Example: 35 The number of solution of 3) log ( )1 log ( 2 4    x x [IIT Screening 2001].

[Audio] Quadratic Equations and Inequations 177 (a) 3 (b) 1 (c) 2 (d) 0 Solution: (b) We have 3) log ( )1 log ( 2 4    x x 3 2) ( )1 (    x x  x x x 6 9 1 2      0 10 7 2  x  x   0 2) 5)( (    x x x  5 or x  2 But x 3  0 , when x  2 .  Only solution is x  5 . Hence number of solution is one. ***.