Quadratic Equations and Inequations PDF

158 158 Quadratic Equations and Inequations 60 4.1 Polynomial. Algebraic expression containing many terms of the form cx n , n being a non-negative variable, a0 , a1 , a2.......an are constants and a n  0 Example : 4 x 4  3 x 3  7 x 2  5 x  3 , 3 x 3  x 2  3 x  5. E3 integer is called a poly...

158 158 Quadratic Equations and Inequations 60 4.1 Polynomial. Algebraic expression containing many terms of the form cx n , n being a non-negative variable, a0 , a1 , a2.......an are constants and a n  0 Example : 4 x 4  3 x 3  7 x 2  5 x  3 , 3 x 3  x 2  3 x  5. E3 integer is called a polynomial. i.e., f (x )  a0  a1 x  a 2 x 2  a 3 x 3 ......  an 1 x n 1  an x n , where x is a (1) Real polynomial : Let a0 , a1 , a2.......an be real numbers and x is a real variable. ID Then f (x )  a0  a1 x  a 2 x 2  a 3 x 3 ......  an x n is called real polynomial of real variable x with real coefficients. Example : 3 x 3  4 x 2  5 x  4 , x 2  2 x  1 etc. are real polynomials. U (2) Complex polynomial : If a0 , a1 , a2.......an be complex numbers and x is a varying complex number. D YG Then f (x )  a0  a1 x  a 2 x 2  a 3 x 3 ......  an x n is called complex polynomial of complex variable x with complex coefficients. Example : 3 x 2  (2  4 i)x  (5 i  4 ), x 3  5 i x 2  (1  2i)x  4 etc. are complex polynomials. (3) Degree of polynomial : Highest power of variable x in a polynomial is called degree of polynomial. Example : f (x )  a0  a1 x  a 2 x 2 ......  an 1 x n 1  an x n is a n degree polynomial. f (x )  4 x 3  3 x 2  7 x  5 is a 3 degree polynomial. U f (x )  3 x  4 is single degree polynomial or linear polynomial. f (x )  bx is an odd linear polynomial. ST A polynomial of second degree is generally called a quadratic polynomial. Polynomials of degree 3 and 4 are known as cubic and biquadratic polynomials respectively. (4) Polynomial equation : If f(x) is a polynomial, real or complex, then f(x) = 0 is called a polynomial equation. 4.2 Types of Quadratic Equation. A quadratic polynomial f(x) when equated to zero is called quadratic equation. Example : 3 x 2  7 x  5  0,  9 x 2  7 x  5  0, x 2  2 x  0, 2 x 2  0 or An equation in which the highest power of the unknown quantity is two is called quadratic equation. Quadratic equations are of two types : 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. ax 2  c  0 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. 60 i.e. ax 2  bx  c  0 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. E3 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 2  3 is a root of given equation, then 2  3 is also its ID 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. D YG 4.3 Solution of Quadratic Equation. U  (1) Factorization method : Let ax 2  bx  c  a(x   )(x   )  0. 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 : 3 x 2  2 x  1  0  (x  1)(3 x  1)  0 x  1,  1 / 3 U (2) Hindu method (Sri Dharacharya method) : By completing the perfect square as ax 2  bx  c  0  x 2  b c x 0 a a ST 2 2 b  b 2  4 ac   b   x   Adding and subtracting   ,  0  2a  4 a 2   2 a   which gives, x   b  b 2  4 ac 2a Hence the quadratic equation ax 2  bx  c  0 (a  0) has two roots, given by   b  b 2  4 ac  b  b 2  4 ac ,   2a 2a Note :  Every quadratic equation has two and only two roots. 4.4 Nature of Roots. 160 Quadratic Equations and Inequations In quadratic equation ax 2  bx  c  0 , the term b 2  4 ac 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 ax 2  bx  c  0 has non-real complex roots. b  D 2a E3  b  D , 2a 60 (ii) If D > 0, then equation ax 2  bx  c  0 has real and distinct roots, namely   and then ax 2  bx  c  a (x   )(x   ) …..(i) (iii) If D = 0, then equation ax 2  bx  c  0 has real and equal roots     and then ax 2  bx  c  a (x   )2 b 2a ID …..(ii) To represent the quadratic expression ax 2  bx  c in form (i) and (ii), transform it into linear factors. U (iv) If D  0, then equation ax 2  bx  c  0 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. D YG (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 U (i) If D1  D 2  0 , then (a) At least one of D1 and D 2  0. (b) If D1  0 then D 2  0 ST (ii) If D1  D 2  0 , then (a) At least one of D1 and D 2  0. (b) If D1  0 then D 2  0. 4.5 Roots Under Particular Conditions. For the quadratic equation ax 2  bx  c  0. (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. Quadratic Equations and Inequations 161 a  0 c  0   roots are of opposite signs. a  0 c  0 (6) If a  0 b  0 c  0   both roots are negative, provided D  0. a  0 b  0 c  0 (7) If a  0 b  0 c  0   both roots are positive, provided D  0. a  0 b  0 c  0 60 (5) If (8) If sign of a = sign of b  sign of c  greater root in magnitude, is negative. E3 (9) If sign of b = sign of c  sign of a  greater root in magnitude, is positive. (10) If a  b  c  0  one root is 1 and second root is c/a. (11) If a  b  c  0 , then equation will become an identity and will be satisfied by every value of x. ID (12) If a  1 and b, c  I and the root of equation ax 2  bx  c  0 are rational numbers, then these roots must be integers. U 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. Both the roots of given equation (x  a)(x  b)  (x  b)(x  c)  (x  c)(x  a)  0 are always D YG Example: 1 (a) Positive Solution: (c) Given equation (b) Negative [MNR 1986; IIT 1980; Kurukshetra CEE 1998] (c) Real (x  a)(x  b)  (x  b)(x  c)  (x  c)(x  a)  0 (d) Imaginary can be re-written as 3 x  2(a  b  c)x  (ab  bc  ca)  0 2 D  4[(a  b  c)2  3(ab  bc  ca)]  4[a2  b 2  c 2  ab  bc  ac]  2[(a  b)2  (b  c)2  (c  a)2 ]  0 Hence both roots are always real. If the roots of (b  c)x 2  (c  a)x  (a  b)  0 are equal then a  c  U Example: 2 (a) 2b Solution: (a) (b) b 2 (c) 3b [Kurukshetra CEE 1992] (d) b b c c a a b  0 ST Hence one root is 1. Also as roots are equal, other root will also be equal to 1. Also .   Example: 3 ab ab  1.1   a  b  b  c  2b  a  c b c b c If the roots of equation (a) 2r Solution: (a) 1 1 1 are equal in magnitude but opposite in sign, then ( p  q )    x  p x q r [Rajasthan PET 1999] (b) r (c) – 2r (d) None of these Given equation can be written as x  (p  q  2r)x  [ pq  (p  q)r]  0 2 Since the roots are equal and of opposite sign,  Sum of roots = 0  ( p  q  2r)  0  p  q  2r Example: 4 If 3 is a root of x 2  kx  24  0 , it is also a root of (a) x  5 x  k  0 2 (b) x  5 x  k  0 2 [EAMCET 2002] (c) x  kx  6  0 2 (d) x  kx  24  0 2 162 Quadratic Equations and Inequations Equation x 2  kx  24  0 has one root as 3, Solution: (c)  3 2  3k  24  0  k  5 Put x  3 and k  5 in option Only (c) gives the correct answer i.e.  3 2  15  9  0  0  0 For what values of k will the equation x 2  2(1  3k )x  7(3  2k )  0 have equal roots (a) 1, –10/9 (b) 2, –10/9 (c) 3, –10/9 [MP PET 1997] (d) 4, –10/9 60 Example: 5 Since roots are equal then [2(1  3k )]  4.1.7(3  2k )  1  9 k  6 k  21  14 k  9 k 2  8 k  20  0 2 2 Solution: (b) Solving, we get k  2,  10 / 9 4.6 Relations between Roots and Coefficients. E3 (1) Relation between roots and coefficients of quadratic equation : If  and  are the roots of quadratic equation ax 2  bx  c  0 , (a  0) then b coefficien t of x Sum of roots  S       a coefficien t of x 2 c constant term  a coefficien t of x 2 ID Product of roots  P  .  If roots of quadratic equation ax 2  bx  c  0 (a  0) are  and  then b 2  2ac a2 D YG (ii)  2   2  (   )2  2  b 2  4 ac  D  a a U (i) (   )  (   )2  4   (iii)  2   2  (   ) (   ) 2  4   (iv)  3   3  (   )3  3(   )   b b 2  4 ac  b D  a2 a2 b(b 2  3 ac) a3 U (v)  3   3  (   )3  3 (   )  (   ) 2  4 {(   ) 2   }  (vi)    4 2  b 2  2ac  c2  2  {(   )  2}  2    2  a2  a  2 2 ST 4 2 2 (vii)  4   4  ( 2   2 )( 2   2 )   b(b 2  2 ac) b 2  4 ac a4 (viii)  2     2  (   )2    b 2  ac a2 (ix)    2   2 (   ) 2  2 b 2  2ac         ac (x)  2    2  (   )   2 bc a2   a 4   4 ( 2   2 )2  2 2  2 b 2 D  2a 2 c 2  (xi)          2 2  2 2 a2c 2    2  (b 2  ac) b 2  4 ac a3 Quadratic Equations and Inequations 163 (2) Formation of an equation with given roots : A quadratic equation whose roots are  and  is given by (x   )(x   )  0  x 2  (   )x    0 i.e. x 2  (sum of roots) x  (product of roots )  0  x 2  Sx  P  0 (3) Equation in terms of the roots of another equation : If ,  are roots of the equation 60 ax 2  bx  c  0 , then the equation whose roots are (Replace x by – x) (ii) 1 /  , 1 /   cx 2  bx  a  0 (Replace x by 1/x) (iii)  n ,  n ; n  N  a(x 1 / n )2  b(x 1 / n )  c  0 (Replace x by x 1 / n ) (iv) k, k  ax 2  kbx  k 2 c  0 (Replace x by x/k) (v) k   , k    a(x  k )  b(x  k )  c  0 (Replace x by (x – k)) 2    k 2 ax 2  kbx  c  0 , k k (Replace x by kx) ID (vi) E3 (i) –, –  ax 2  bx  c  0 (vii)  1 / n ,  1 / n ; n  N  a(x n )2  b(x n )  c  0 (Replace x by x n ) U (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 . (i)  2   2 D YG Some examples of symmetric expressions are : (ii)  2     2 2 (v)  2    2    (vi)         (iii) 1   1  (iv)      2 (vii)  3   3 (viii)  4   4 4.7 Biquadratic Equation. ST U If , , ,  are roots of the biquadratic equation ax 4  bx 3  cx 2  dx  e  0 , then c c S 1          b / a , S 2  .  .      .    (1)2  a a d or S 2  (   )(   )      c / a , S 3          (1)3  d / a a e e or S 3  (   )   (   )  d / a and S 4  ...  (1)4  a a Example: 6 If the difference between the corresponding roots of x 2  ax  b  0 and x 2  bx  a  0 is same and a  b , then [AIEEE 2002] (a) a  b  4  0 Solution: (a) (b) a  b  4  0 (d) a  b  4  0     a ,   b      a 2  4 b and     b ,   a      b 2  4 a According to question,         Example: 7 (c) a  b  4  0 a2  4 b  b 2  4 a  a  b  4  0 If the sum of the roots of the quadratic equation ax 2  bx  c  0 is equal to the sum of the squares of their reciprocals, then a / c, b / a, c / b are in [AIEEE 2003; DCE 2000] (a) A.P. (b) G.P. (c) H.P. (d) None of these 164 Quadratic Equations and Inequations Solution: (c) As given, if ,  be the roots of the quadratic equation, then    1 2  1 2  (   )2  2  2 2   b b 2 / a 2  2 c / a b 2  2 ac   a c 2 / a2 c2 2a b c 2a b 2 b ab 2  bc 2  2a 2 c  ab 2  bc 2     2   b c a c a c ac 2 a b c c a b , , are in A.P.  , , are in H.P. a b c c a b 60  Example: 8 Let ,  be the roots of x 2  x  p  0 and ,  be root of x 2  4 x  q  0. If , , ,  are in G.P., then the Solution: (a) integral value of p and q respectively are (a) – 2, – 32 (b) – 2, 3     1 ,   p ,     4 ,   q [IIT Screening 2001] (c) – 6, 3 E3 Since , , ,  are in G.P. r   /   /    / (d) – 6, – 32   r  1   (1  r)  1 ,  (r 2  r 3 )  4  .r 2 (1  r)  4 ID So r 2  4  r  2 If r  2 ,   2  1    1 / 3 and r  2 ,   2  1    1 But p    I  r  2,   1  p  2 , q   2r 5  1(2)5  32 If 1 – i is a root of the equation x 2  ax  b  0 , then the values of a and b are (a) 2, 1 Solution: (b) U Example: 9 (b) – 2, 2 (c) 2, 2 [Tamil Nadu Engg. 2002] (d) 2, – 2 Since 1  i is a root of x  ax  b  0.  1  i is also a root. 2 D YG Sum of roots  1  i  1  i  a  a  2 Product of roots  (1  i)(1  i)  b  b  2 Hence a  2 , b  2 Example: 10 If the roots of the equation x 2  5 x  16  0 are ,  and the roots of equation x 2  px  q  0 are  2   2 ,  / 2 , then (a) p  1, q  56 Solution: (b) (b) p  1, q  56 [MP PET 2001] (c) p  1, q  56 (d) p  1, q  56 Since roots of the equation x 2  5 x  16  0 are  , . U      5,   16 and  2   2   2   p  (   )2  2   2   p  25  2(16 )  16   p  p  1 2 ST      q  (25  32 )8  q  q  56 and ( 2   2 )    q  [(   )2  2 ] 2  2  Example: 11 If    , but  2  5  3,  2  5   3 , then the equation whose roots are [EAMCET 1989; AIEEE 2002] (a) x  5 x  3  0 2 Solution: (b) S S  S  (b) 3 x  19 x  3  0 2    2   2 5  3  5   3        5(   )  6    and is   , p 5(5)  6 19  3 3 (c) 3 x  12 x  3  0 2 (d) None of these   2  5  3      2  5   3      1  p  1. ,  are roots of x 2  5 x  3  0. Therefore     5 ,   3   Quadratic Equations and Inequations 165 ∵ x2  Example: 12 19 x  1  0  3 x 2  19 x  3  0 3 Let ,  be the roots of the equation (x  a)(x  b)  c , c  0 , then the roots of the equation (x   )(x   )  c  0 are [IIT 1992; DCE 1998, 2000; Roorkee 2000] (a) a, c (b) b, c (c) a, b (d) a, d Since ,  are the roots of (x  a)(x  b)  c i.e. of x 2  (a  b)x  ab  c  0      a  b  a  b     and   ab  c  ab    c  a, b are the roots of x 2  (   )x    c  0  (x   )(x   )  c  0 Hence (c) is the correct answer If  and  are roots of the equation x 2  ax  b  0 and Vn   n   n , then [Rajasthan PET 1995; Karnataka CET 200 (c) Vn 1  a Vn  b Vn 1 (d) Vn 1  b Vn  a Vn 1 (b) Vn 1  b Vn  a Vn 1 (a) Vn 1  a Vn  b Vn 1 Solution: (a) E3 Example: 13 60 Solution: (c) Since  and  are roots of equation, x  ax  b  0 , therefore     a ,   b 2 Now, Vn 1   n 1   n 1  (   )( n   n )   ( n 1   n 1 )  Vn 1  a. Vn  b. Vn 1 If one root of the equation x 2  px  q  0 is the square of the other, then (b) p  q  q(1  3 p)  0 (a) p  q  q(3 p  1)  0 3 3 2 2 Let  and  2 be the roots then    2   p , . 2  q U Solution: (d) 2 (d) p 3  q 2  q(1  3 p)  0 (c) p  q  q(3 p  1)  0 3 [IIT Screening 2004] ID Example: 14 Now (   2 )3   3   6  3 3 (   2 )   p 3  q  q 2  3 pq  p 3  q 2  q(1  3 p)  0 Example: 15 Let  and  be the roots of the equation x 2  x  1  0 , the equation whose roots are  19 ,  7 is[IIT 1994; Pb. CET D YG Solution: (d) (c) x 2  x  1  0 (b) x 2  x  1  0 (a) x 2  x  1  0 Roots of x 2  x  1  0 are x  (d) x 2  x  1  0 1  1  4  1  3i ,  ,  2 2 2 Take    ,    2   19  w19  w,  7  (w 2 )7  w14  w 2  Required equation is x 2  x  1  0 Example: 16 If one root of a quadratic equation is U Given root  1 2 5 , then the equation is  [Rajasthan PET 1987] (c) x 2  4 x  1  0 (b) x 2  4 x  1  0 (a) x 2  4 x  1  0 Solution: (b) 1 2 5 (d) None of these 2 5  2  5 ,  other root  2  5 1 ST Again, sum of roots = – 4 and product of roots = – 1. The required equation is x 2  4 x  1  0 4.8 Condition for Common Roots. (1) Only one root is common : Let  be the common root of quadratic equations a1 x  b1 x  c1  0 and a 2 x 2  b 2 x  c 2  0. 2  a1 2  b1  c1  0 , a2 2  b2  c2  0 By Crammer’s rule :  2  c1  c2 b1 b2  a2 c1  a1c2 b1c2  b2 c1  ,  0 a1b2  a2b1 a2c1  a1c2  a1 a2  c1  c2  1 a1 a2 b1 b2 or 2 b1c 2  b 2 c1   a2 c1  a1c 2  1 a1b 2  a2b1 166 Quadratic Equations and Inequations  The condition for only one root common is (c1 a 2  c 2 a1 )2  (b1 c 2  b 2 c1 )(a1 b 2  a 2 b1 ) (2) Both roots are common: Then required condition is a1 b1 c1.   a2 b 2 c 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. If one of the roots of the equation x 2  ax  b  0 and x 2  bx  a  0 is coincident. Then the numerical value of (a  b ) is E3 Example: 17 60  [IIT 1986; Rajasthan PET 1992; EAMCET 2002] (a) 0 (c) (d) 5 If  is the coincident root, then   a  b  0 and   b  a  0 2  2 a b 2  2  b a  1 b a 2 ID Solution: (b) (b) – 1   (a  b) ,   1  (a  b)  1  (a  b)  1 2 Example: 18 If a, b, c are in G.P. then the equations ax 2  2bx  c  0 and dx 2  2ex  f  0 have a common root if U d e f , , are in a b c Solution: (a) (b) G.P. D YG (a) A.P. [IIT 1985; Pb. CET 2000; DCE 2000] (c) H.P. (d) None of these As given, b 2  ac  ax 2  2bx  c  0 can be written as ax 2  2 ac x  c  0  ( a x  c )2  0  x   c a This must be common root by hypothesis c c f 0 So it must satisfy the equation, dx 2  2ex  f  0  d    2e a a   d f 2e   a c c 2e d f 2e   a c b  c. a d e f , , are in A.P. a b c U Hence c  a 4.9 Properties of Quadratic Equation. ST (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 f (x )  0 lie between a and b. Y O Y f(a) = +ve f(a) = +ve x=b A x=a x=a X x=b X B f(b) = – ve f(b) = – ve (2) If f (a)  f (b ) then there exists a point c between a and b such that f (c)  0 , a  c  b. Y Q P O a A b B X Quadratic Equations and Inequations 167 60 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. f (x )  0 at x  c. (3) If  is a root of the equation f (x )  0 then the polynomial f (x ) is exactly divisible by (x   ) or (x   ) is factor of f (x ).    ratio  i.e. 1  2 1  2  E3 (4) If the roots of the quadratic equations ax 2  bx  c  0 , a 2 x 2  b 2 x  c 2  0 are in the same   then b12 / b 22  a1 c1 / a 2 c 2.  ID (5) If one root is k times the other root of the quadratic equation ax 2  bx  c  0 then (k  1) 2 b 2 . k ac The value of ‘a’ for which one root of the quadratic equation (a 2  5 a  3) x 2  (3 a  1)x  2  0 is twice as large as the other is [AIEEE 2003] (a) 2/3 (b) – 2/3 (c) 1/3 (d) – 1/3 Let the roots are  and 2 1  3a 1  3a 2 2 Now,   2  2 , . 2  2  3  2 , 2 2  2 a  5a  3 a  5a  3 a  5a  3 a  5a  3 D YG Solution: (a) U Example: 19  1 (1  3 a)2  2 (1  3 a)2  2  2  2  9  9a2  45 a  27  1  9a2  6a  39 a  26  a  2 / 3 2 2 a  5a  3  9 (a  5 a  3)  a  5 a  3 4.10 Quadratic Expression. An expression of the form ax 2  bx  c , where a, b, c  R and a  0 is called a quadratic U expression in x. So in general, quadratic expression is represented by f (x )  ax 2  bx  c or y  ax 2  bx  c. (1) Graph of a quadratic expression : We have y  ax 2  bx  c  f (x ) ST 2 2  D b  b  D    a x  y  a  x     2  y 4a 2a  2a  4 a    b D  Y and X  x  2a 4a 1 Y  a.X 2  X 2  Y a (i) The graph of the curve y  f (x ) is parabolic. Now, let y  b  0 i.e. (parallel to y-axis). 2a (iii) (a) If a > 0, then the parabola opens upward. (ii) The axis of parabola is X  0 or x  168 Quadratic Equations and Inequations (b) If a < 0, then the parabola opens downward. x axis a > 0, D < 0 a < 0, D < 0 60 x axis (iv) Intersection with axis b  D 2a E3 (a) x-axis: For x axis, y  0  ax 2  bx  c  0  x  For D > 0, parabola cuts x-axis in two real and distinct points i.e. x  b  D. 2a ID For D = 0, parabola touches x-axis in one point, x  b / 2a. D> 0 xaxis D YG a>0 a>0 D= 0 D> 0 xaxis U a 0, f(x) has least value at x   E3 (ii) Graphical method : Vertex of the parabola Y  aX 2 is X  0 , Y  0 D b i.e., x   0  x  b / 2a , y   D / 4 a 0, y 4a 2a b. This least value is given 2a ID D  b  by f  .  4a  2a  (b) For a < 0, f(x) has greatest value at x  b / 2a. This greatest value is U D  b  given by f  .  4a  2a  a>0 verte x verte x a 0 and D < 0, so f (x )  0 for all x  R i.e., f (x ) is positive for all real values of x. (ii) a < 0 and D < 0, so f (x )  0 for all x  R i.e., f(x) is negative for all real values of x. (iii) a > 0 and D = 0 so, f (x )  0 for all x  R i.e., f(x) is positive for all real values of x except at vertex, where f (x )  0. (iv) a < 0 and D = 0 so, f (x )  0 for all x  R i.e. f(x) is negative for all real values of x except at vertex, where f (x )  0. U (v) a > 0 and D > 0 Let f (x )  0 have two real roots  and  (   ) , then f (x )  0 for all x  (,  )  ( , ) and ST f (x )  0 for all x  ( ,  ). (vi) a  0 and D  0 Let f (x )  0 have two real roots  and  (   ) , Then f (x )  0 for all x  (,  )  ( , ) and f (x )  0 for all x  ( ,  ) Example: 20 If x be real, then the minimum value of x 2  8 x  17 is (a) – 1 (b) 0 (c) 1 Solution: (c) Since a  1  0 therefore its minimum value is  Example: 21 If x is real, then greatest and least values of (a) 3, –1/2 (b) 3, 1/3 [MNR 1980] (d) 2 4 ac  b 4 (1)(17 )  64 4   1 4a 4 4 2 x2  x 1 are x2  x 1 (c) – 3, –1/3 [IIT 1968; Rajasthan PET 1988] (d) None of these 170 Quadratic Equations and Inequations Solution: (b) Let y  x2  x 1 x2  x 1 x 2 (y  1)  (y  1)x  (y  1)  0 ∵ x is real, therefore b 2  4 ac  0 60 1 1   (y  1)2  4(y  1)(y  1)  0  3 y 2  10 y  3  0  (3 y  1)(y  3)  0   y   (y  3)  0   y  3 3 3  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 g(x )  f (x )  f (x )  f  (x ). Then for any real value of x Solution: (b) (c) g(x )  0 (b) g(x )  0 (d) g(x )  0 E3 (a) g(x )  0 [IIT 1990] Let f (x )  ax  bx  c , then g(x )  ax  bx  c  2ax  b  2a  ax  (b  2a)x  (b  c  2a) 2 2 2 ∵ f ( x )  0. Therefore b 2  4 ac  0 and a  0 Now for g(x), Discriminant  (b  2a)2  4 a(b  c  2a)  b 2  4 a2  4 ab  4 ab  4 ac  8 a 2  (b 2  4 ac)  4 a2  0 as b 2  4 ac  0 If ,  (   ) are roots of the equation x 2  bx  c  0 where (c  0  b ) then (a) 0     Solution: (b) (b)   0   |  | Since f (0)  0  0  c  c  0 (c)     0 [IIT Screening 2000] (d)   0 |  |   U Example: 23 ID Therefore sign of g(x) and a are same i.e. g(x )  0.  Roots will be of opposite sign,     b  ve (b > 0) D YG It is given that    So,     ve is possible only when |  |      0,   0,|  |      0   |  | 4.11 Wavy Curve Method. Let f (x )  (x  a1 )k1 (x  a 2 )k 2 (x  a 3 )k 3.......( x  an 1 )k n 1 (x  an )k n Where k 1 , k 2 , k 3..., k n  N and U condition …..(i) a1 , a 2 , a 3 ,......, a n are fixed natural numbers satisfying the a1  a 2  a 3.....  a n 1  an First we mark the numbers a1 , a 2 , a 3 ,......, a n on the real axis and the plus sign in the interval ST of the right of the largest of these numbers, i.e. on the right of a n. If k n is even then we put plus sign on the left of a n and if k n is odd then we put minus sign on the left of a n. In the next interval we put a sign according to the following rule : When passing through the point a n 1 the polynomial f(x) changes sign if k n 1 is an odd number and the polynomial f(x) has same sign if k n 1 is an even number. Then, we consider the next interval and put a sign in it using the same rule. Thus, we consider all the intervals. The solution of f (x )  0 is the union of all intervals in which we have put the plus sign and the solution of f (x )  0 is the union of all intervals in which we have put the minus sign. 4.12 Position of Roots of a Quadratic Equation. Quadratic Equations and Inequations 171 Let f (x )  ax 2  bx  c , where a, b, c  R be a quadratic expression and k , k 1 , k 2 be real numbers such that k 1  k 2. Let ,  be the roots of the equation f (x )  0 i.e. ax 2  bx  c  0. Then  b  D b  D ,   where D is the discriminant of the equation. 2a 2a a>0 (–b/2a, – D/4a) f(k )  k  –  k  xaxis f(k ) E3 – b/2a 60 (1) Condition for a number k (If both the roots of f(x) = 0 are less than k) xaxis b/2a a0 f(k ) k  –  xaxis b/2a f(k ) a0 (iii) k  b / 2a , (–b/2a, – U (3) Condition for a number k (If k lies between the roots of f(x)D/4a) = 0) ST  k  xaxis  k (–b/2a, – D/4a)  xaxis a0 k1  f(k (k2 2) ) k1  xaxis x axis f(k1 ) f(k 2) (k 2 ) a0 k1  f(k 2)  k1  k2  E3 f(k1 ) 60 (5) Condition for numbers k1 and k2 (If both roots of f(x) = 0 are confined between k1 and k2) k2 a0  (i) D  0 k1  k1 k2  x-axis k2 f(k1 f(k ) 2)  a