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

⇒ D = 4(k – 2)

For equal roots, D = 0

Thus, 4(k – 2) = 0

So, k = 2.