Find the values of p for which the quadratic equation (p + 1)x^{2} – 6 (p + 1) x + 3 (p + 9) = 0, p 1 has equal roots. Hence, find the roots of the equation.

Given equation is (p + 1)x^{2} – 6 (p + 1) x + 3 (p + 9) = 0

Comparing with standard quadratic equation ax^{2} + bx + c = 0

a = (p + 1) b = – 6(p + 1) c = 3(p + 9)

Given that the roots of equation are equal

Thus D = 0

Discriminant D = b^{2} – 4ac = 0

[ – 6(p + 1)]^{2} – 4.(p + 1).3(p + 9) = 0

36(p + 1)(p + 1) – 12(p + 1)(p + 9) = 0

12(p + 1)[3(p + 1) – (p + 9)] = 0

12(p + 1)[3p + 3 – p – 9] = 0

12(p + 1)[2p – 6] = 0

(p + 1) = 0 or [2p – 6] = 0

p = – 1 or p = 3

The values of p are – 1, 3 for which roots of the quadratic equation are real and equal.

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