Question

Find the value of k for which the equation x² + k(2x + k − 1) + 2 = 0 has real and equal roots.

Answer 1

The given equation is x²+k(2x+k−1)+2=0.

⇒x²+2kx+k(k−1)+2=0

So, a = 1, b = 2k, c = k(k − 1) + 2

We know D=b²−4ac

⇒D=(2k)² − 4 × 1 × [k(k − 1) + 2]

⇒D=4k² − 4[k² − k + 2]

⇒D=4k² − 4k² + 4k − 8

⇒D=4k − 8 = 4(k − 2)

For equal roots, D = 0

Thus, 4(k − 2) = 0

So, k = 2.