Question

Find the value of 'p' for which the quadratic equation p(x – 4)(x – 2) + (x –1)² = 0 has real and equal roots.

Answer 1

Given quadratic equation is

p(x – 4)(x – 2) + (x –1)² = 0

⇒ p(x² – 4x – 2x + 8) + (x² + 1 – 2x) = 0

⇒ px² – 6px + 8p + x² + 1 – 2x = 0

⇒ x²(p + 1) – 2x(3p + 1) + (8p + 1) = 0

Comparing the above equation with ax² + bx + c = 0, we get

a = p + 1, b = –2(3p + 1) and c = 8p + 1

For real and equal roots

D = 0 i.e., b² – 4ac = 0

∴ [–2(3p + 1)]² – 4(p + 1)(8p + 1) = 0

⇒ 4(3p + 1)² – 4(8p² + 9p + 1) = 0

⇒ 4(9p² + 1 + 6p) – 32p² – 36p – 4 = 0

⇒ 36p² + 4 + 24p – 32p² – 36p – 4 = 0

⇒ 4p² – 12p = 0

⇒ 4p(p – 3) = 0

⇒ p = 0 or p = 3

Hence, for p = 0 or p = 3, the given quadratic equation has real and equal roots.