Question

Find the value of k for which x = 3 is a solution of the quadratic equation (k + 2)x² – kx + 6 = 0. Thus, find the other root of the equation.

Answer 1

Substituting, x = 3 in (k + 2)x² – kx + 6 = 0 we get,

⇒ (k + 2)(3)² – 3k + 6 = 0

⇒ (k + 2)(9) – 3k + 6 = 0

⇒ 9k + 18 – 3k + 6 = 0

⇒ 6k + 24 = 0

⇒ 6k = –24

⇒ k = `(-24)/6`

⇒ k = –4

Substitute the value of k = –4 in (k + 2)x² – kx + 6 = 0 we get,

⇒ (–4 + 2)(x)² – (–4)x + 6 = 0

⇒ (–2)(x)² – (–4)x + 6 = 0

⇒ –2x² + 4x + 6 = 0

⇒ –2x² – 2x + 6x + 6 = 0

⇒ –2x(x + 1) + 6(x + 1) = 0

⇒ (x + 1)(–2x + 6) = 0

⇒ (x + 1) = 0 or (–2x + 6) = 0   ...[Using zero-product rule]

⇒ x = –1 or –2x = –6

⇒ x = –1 or x = `(-6)/(-2)`

⇒ x = –1 or x = 3

Hence, the value of k = –4 and the other root is –1.