Quadratic Equation– Session 3

Session Objective 1. Condition for common root 2. Set of solution of quadratic inequation 3. Cubic equation

Condition for Common Root The equations ax 2 + bx + c = 0 & a’x 2 + b’x + c’ = 0 has a common root(CR)  a  2 + b  + c = 0 a’  2 + b’  + c’ = 0 By rule of cross- multiplication Treating  2 and  as two different variable bc’–cb’ca’–ac’ ab’–ba’ _H007

Condition for Common Root Condition for common root of ax 2 + bx + c = 0 & a’x 2 + b’x + c’ = 0 is (ca’-ac’) 2 =(bc’-cb’)(ab’-ba’) _H007

Illustrative Problem If x 2 -ax-21=0 and x 2 -3ax+35=0 (a>0)has a common root then value of a is (a)3 (b) 4 (c )2 (d) 4 _H007

Illustrative Problem Let  be the common root  2 -a -21=0  2 -3a+35=0 By Cross- Multiplication Solution: Method 1 If x 2 -ax-21=0 and x 2 -3ax+35=0 (a>0)has a common root then value of a is (a)3 (b)4 (c )2 (d)4 _H007

Illustrative Problem a 2 =16 a =  4 As a>0  a=4 If x 2 -ax-21=0 and x 2 -3ax+35=0 (a>0)has a common root then value of a is (a)3 (b)  4 (c )2 (d) 4 _H007

Illustrative Problem If x 2 -ax-21=0 and x 2 -3ax+35=0 (a>0)has a common root then value of a is (a)3 (b)  4 (c )2 (d) 4  2 - a - 21 = 0 ….(A)  2 -3a+35 = 0 …..(B) (A) – (B)  2a = 56 Substituting ‘’ in (A) a =  4 As a>0  a=4 Solution: Method 2 _H007

Illustrative Problem If equation x 2 -ax+b=0 and x 2 +bx-a=0 has only one common root then prove that a-b=1 Solution: x 2 - ax + b=0 …(A) x 2 + bx - a=0 …(B) By observation at x=1 both the equation gives same value. L.H.S. = a-b-1 for x=1 This means x=1 is the common root  a – b – 1= 0  a–b=1 Why? for x=1 both the equations give this _H007

Illustrative Problem If equation x 2 -ax+b=0 and x 2 +bx-a=0 has only one common root then prove that a-b=1 Solution: Let  be the common root then  2 -a + b = 0 &  2 + b - a = 0 subtracting one from the other we get (b + a)  - (b + a) = 0  = 1 provided b + a   0 Hence x = 1 is the common root  1 – a + b = 0 or a – b = 1 _H007 Method 2 Why??

Condition for Two Common Roots The equations ax 2 +bx+c=0 and a’x 2 +b’x+c’ = 0 have both roots common For two roots to be common ax 2 + bx + c  K(a’x 2 + b’x + c’) why? when both the roots are common,two equations will be same. But not necessarily identical. As x 2 –3x+2=0 and 2x 2 –6x+4=0 Same equation. Both have roots 1,2 But not identical _H007

Illustrative Problem If x 2 +ax+(a-2) = 0 and bx 2 +2x+ 6 = 0 have both roots common then a : b is (a) 2 (b)1/2 (c) 4(d)1/4 Solution: As both roots are common  a=-1 _H007

Quadratic Inequation If ax 2 +bx+c =0 has roots ,; let < ax 2 +bx+c = a(x- )(x- ) ax 2 +bx+c > 0 ax 2 +bx+c  0 A statement of inequality exist between L.H.S and R.H.S Quadratic Inequation When ax 2 +bx+c >0Let a>0  (x- )(x- )>0 _H009

Quadratic Inequation (x- )(x- )>0 Either (x- )>0; (x- )>0 Or (x- )<0; (x- )<0  x >   x>> x>  x <   x<< x< for a(x- )(x- )>0 ; (a>0) x lies outside , Number line and x>  and x<   and  are not included in set of solutions   -- _H009

Illustrative Problem Find x for which 6x 2 -5x+1>0 holds true  Either x>1/2 or x<1/3 1/31/2  -- for 6 (x-1/3) (x-1/2)>0 Solution: _H009

Quadratic Inequation When ax 2 +bx+c < 0  (x- )(x- ) < 0 ax 2 +bx+c = a(x- )(x- ) and a>0 Either (x- ) 0 Or (x- )>0; (x- )<0  x <   x >   <x< where  <  and x>  No solution and x<  for a(x- )(x- ) 0) x lies within ,  and  are not included in set of solutions <x<   -- _H009

Illustrative Problem 1/31/2 x  1/3 <x< 1/2 Find x for which 6x 2 -5x+1<0 holds true for 6 (x-1/3) (x-1/2)<0 Solution: _H009

Illustrative Problem Solve for x : x 2 - x – 6 > 0 (x-3) (x+2) >0 Step1:factorize into linear terms -2 3 Step2 :Plot x for which x 2 -x–6=0 on number line As sign of a >0 x 2 -x–6 >0 for either x 3  -- Solution: _H009

Quadratic Inequation For a(x- )(x- )  0  x x   x   Here set of solution contains , and all values outside , For a(x- )(x- )  0  x   x   x lies within , and also includes , in solution set _H009

Illustrative Problem Solution : x 2 – 8x + 10  0 step1: Find the roots of the corresponding equation Roots of x 2 – 8x + 10 = 0 are _H009

Illustrative Problem Step2: Plot on number-line 4-6 and 4+6 are included in the solution set x 2 – 8x + 10  0 4-64+6  -- _H009

Illustrative Problem Solve for x : - x x – 6 > 0 Here a=-1Solution: Step1: Multiply the inequation with (-1)to make ‘a’ positive. Note- Corresponding sign of inequality will also change x 2 –15 x + 6 < 0  (x-7)(x-8) <0 78  -- step2: Plot on Number line 7 <x<8 _H009

Cubic Equation P(x)=ax 3 +bx 2 +cx+d A polynomial of degree 3 P(x)=0  ax 3 +bx 2 +cx+d=0 is a cubic equation when a  0 Number of roots of a cubic equation? _H015

Cubic Equation Let the roots of ax 3 +bx 2 +cx+d =0 be ,,  ax 3 +bx 2 +cx+d a(x- ) (x-  ) (x- ) As ax 2 +bx+c has roots , can be written as ax 2 +bx+c a(x- ) (x-  ) a[x 3 -(++)x 2 +(++)x-()] _H015

Cubic Equation Comparing co-efficient ax 3 +bx 2 +cx+d a[x 3 -(++)x 2 + (++)x-()] _H015

Cubic Equation ax 3 +bx 2 +cx+d=0 a,b,c,dR Maximum real root = ? 3 As degree of equation is 3 Minimum real root? 0? Complex root occur in conjugate pair when co-efficient are real Maximum no of complex roots=2 Minimum no. of real root is 1 _H015

Illustrative Problem If the roots of the equation x 3 -2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is (a)5 (b) –5 (c )4 (d) None of these _H015

Illustrative Problem x 3 -2x+4=0 has roots ,,  a=1, b=0, c=-2, d=4 (1+ ) (1+ ) (1+ )= 1+ + + = = -5 If the roots of the equation x 3 -2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is (a)5 (b)–5 (c )4 (d) None of these Solution Method 1: _H015

Illustrative Problem Method 2 Let f(x)= x 3 -2x+4 = (x- ) (x- )(x- ) for x=-1 f(-1)= 5 = (-1- ) (-1- )(-1- ) (-1) 3 (1+ ) (1+ )(1+ ) = 5  (1+ ) (1+ )(1+ ) = -5 If the roots of the equation x 3 -2x+4=0 are ,, then the value of (1+ ) (1+ ) (1+ ) is (a)5 (b) –5 (c )4 (d) None of these _H015

Class -Exercise

Class Exercise1 If the equations ax 2 + bx + c = 0 and cx 2 + bx + a = 0 have one root common then (a)a + b + c = 0 (b)a + b – c = 0 (c)a – b + c = 0 (d)both (a) or (c) Solution: By observation roots of one equation is reciprocal to other. So both equation will have common root if it becomes 1 or –1

Class Exercise1 If the equations ax 2 + bx + c = 0 and cx 2 + bx + a = 0 have one root common then (a)a + b + c = 0 (b)a + b – c = 0 (c)a – b + c = 0 (d)both (a) and (c) When 1 is common root,a + b + c = 0. when –1 is common root, a – b + c = 0

Class Exercise2 If x 2 + ax + 3 = 0 and bx 2 + 2x + 6 = 0 have both roots common then a : b is (a) 2 (b)1/2 (c) 4(d)1/4 Solution: As both roots are common

Class Exercise3 Solution : x 2 – 10x + 22  0 Factorize into linear terms by using perfect square method

Class Exercise3 5-35+3 x Plot on number-line 5-3 and 5-3 are included in the solution set

Class Exercise4 The number of integral values of x for which (x – 6) (x + 1) < 2 (x – 9) holds true are (a) Two (b)Three (c) One (d) Zero (x – 6) (x + 1) < 2 (x – 9) or, x 2 – 5x – 6 < 2x – 18  x 2 – 7x + 12 < 0 or, (x – 3) (x – 4) < 0  3 < x < 4 So no integral values of x satisfies it. Solution:

Class Exercise5 If , ,  are the roots of 2x 3 + 3x 2 – 2x + 1 = 0. Then the value of is (a)17/4 (b) 41/4 (c )9/4 (d) None of these Solution: 2x 3 + 3x 2 – 2x + 1 = 0

Class Exercise5 If , ,  are the roots of 2x 3 + 3x 2 – 2x + 1 = 0. Then the value of is (a)17/4 (b) 41/4 (c )9/4(d) None of these

Class Exercise6 If ax 2 + bx + c = 0 & bx 2 + c x + a = 0 has a common root and a ‡ 0 then prove that a 3 +b 3 +c 3 =3abc Solution:  (bc – a 2 ) 2 = (ab – c 2 ) (ac – b 2 )

Class Exercise6 If ax 2 + bx + c = 0 & bx 2 + c x + a = 0 has a common root and a ‡ 0 then prove that a 3 +b 3 +c 3 =3abc  (bc – a 2 ) 2 = (ab – c 2 ) (ac – b 2 )  b 2 c 2 + a 4 – 2a 2 bc = a 2 bc – ab 3 – ac 3 + b 2 c 2 By expansion:  a(a 3 + b 3 + c 3 ) = 3a 2 bc a(a 3 + b 3 + c 3 – 3abc) = 0 either a = 0 or a 3 + b 3 + c 3 = 3abc As a  0 a 3 + b 3 + c 3 = 3abc

Class Exercise7 Find the cubic equation with real co-efficient whose two roots are given as 1 and (1 + i) Solution: Imaginary roots occur in conjugate pair; when co-efficients are real Roots are 1, (1 – i) (1 + i) Equation is or, (x – 1) (x 2 – 2x + 2) = 0 x 3 – 3x 2 + 4x – 2 = 0

Class Exercise8 If ax 3 + bx 2 + cx + d = 0 has roots , and  and, then find the equation whose roots are Solution: As roots are incremented by 2. So desired equation can be found replacing x by (x– 2) a(x – 2) 3 + b(x – 2) 2 + c(x – 2) + d = 0

Class Exercise9 Find values of x for which the inequation (2x – 1) (x – 2) > (x – 3) (x – 4) holds true Solution: (2x – 1) (x – 2) >(x – 3) (x – 4)  2x 2 – 5x + 2 > x 2 – 7x + 12  x 2 + 2x – 10 > 0

Class Exercise9 Find values of x for which the inequation (2x – 1) (x – 2) > (x – 3) (x – 4) holds true

Class Exercise10 For what values of ‘a’, a(x-1)(x-2)>0 when 1 < x < 2 (a) a > 0(b) a < 0 (c) a = 0(d) a = 1 Solution: When 1 < x < 2 (x – 1) > 0, (x – 2) < 0 As a (x – 1) (x – 2) > 0 a < 0

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