Question

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

Answer 1

p = +3

Roots of the equation are equal, x = 3

Step-by-step explanation:

Given: (p+1)x² - 6(p+1)x + 3(p + 9) = 0 has equal roots means the discriminant is zero.

p≠-1

d = b² - 4ac = 0

6²(p+1)² - 4[3(p+9)(p+1)] = 0

36(p² + 1 + 2p) - 12 (p² +10p +9) = 0

36p² - 12p² + 72p - 120p +36 - 108 = 0

24p² -48p - 72 = 0

Dividing by 24, we get: p² -2p - 3 = 0

p² -3p +p - 3 = 0

p (p-3) +1 (p-3) = 0

(p-3) (p+1) = 0

p can either be -1 or +3.

Given that p≠-1, we get p = +3.

So the equation becomes:

(3+1)x² - 6(3+1)x + 3(3 + 9) = 0

4x² - 24x +36 = 0

Dividing by 4, we get: x² -6x +9 = 0

x² -3x -3x +9 = 0

x (x -3) -3(x -3) = 0

(x-3)(x-3) = 0

So x = +3 is the roots of the equation