Question

Find the value of k which the roots are real and equal in each of the following equation 1. 5x2-4x+2+k(4x2-2x-1) =0​

Answer 1

Step-by-step explanation:

5x^2 - 4x + 2+ k(4x^2 -2x -1) = 0

5x^2-4x +2+ 4kx^2 -2xk -k =0

(5+4k)x^2-(4+2k)x+(2-k) =0

a = 5+4k b = -(4+2k) c = 2-k

given that roots are real and equal

so, D = 0

D = b^2-4ac

0 = {-(4+2k)}^2 - 4x(5+4k)(2-k)

0 = (16+4k^2+16k)- 4 x (10+3k-4k^2)

0 = 4k^2 +16k +16 -40 -12k + 16k^2

0 = 20k^2 +4k -24

take common 4 from whole eqn

0 = 5k^2+k - 6

it is quadratic equation

solved by factorisation method