Question

the roots of (x+a) (x-b)-8k = (k-2)^2are real and equal where a,b,c € R, then:​

Answer 1

x+a)( x-b) - 8 k = ( k-2)^2

x^2 - bx + ax - ab - 8 k - ( k-2)^2= 0

x^2 +x( a- b) - ab - 8 k -( k-2)^2= 0

As roots are real and equal

So discriminant = 0

( a-b)^2 = 4( - ab - 8 k - ( k-2)^2)

a^2 + b^2 - 2ab = - 4ab - 32 k

- 4( k-2)^2

a^2 + b^2 - 2ab + 4ab = -32k - 4 k^2 -16

+ 16k

a^2 + b^2 + 2ab = - 4 k^2 - 16 k -16

( a+ b)^2 = -4( k^2 + 4k +4)

( a+ b)^2 = -4( k+2)^2

its possible only when

a+ b= k+2 = 0

a= -b, k= -2