Question

Find that value of p for which the quadratic equation (p + 1)x² – 6(p + 1)x + 3(p + 9) = 0, p ≠ –1 has equal roots. Hence, find the roots of the equation.

Answer 1

It is given that the quadratic equation (p + 1)x² – 6(p + 1)x + 3(p + 9) = 0, p ≠ –1 has equal roots.

Therefore, the discriminant of the quadratic equation is 0.

Here, 

a = (p + 1)

b = –6(p + 1)

c = 3(p + 9)

∴ D = b² – 4ac = 0

⇒ [–6(p + 1)]² – 4 × (p + 1) × 3(p + 9) = 0

⇒ 36(p + 1)² – 12(p + 1)(p + 9) = 0

⇒ 12(p + 1)[3(p + 1) – (p + 9)] = 0

⇒ 12(p + 1)(2p – 6) = 0

⇒ p + 1 = 0 or 2p – 6 = 0

⇒ p + 1 = 0

⇒ p = –1

This is not possible as p ≠ –1

2p – 6 = 0

⇒ p = 3

So, the value of p is 3.

Putting p = 3 in the given quadratic equation, we get

(3 + 1)x² – 6(3 + 1)x + 3(3 + 9) = 0

⇒ 4x² – 24x + 36 = 0

⇒ 4(x² – 6x + 9) = 0

⇒ (x – 3)² = 0

⇒ x – 3 = 0

⇒ x = 3

Thus, the root of the given quadratic equation is 3.