Question

Find the value of ‘p’ for which the roots of the following equation are real and equal:

(p + 1) x² − 2(p − 1) x + 1 = 0

Answer 1

Given:

(p + 1) x² − 2(p − 1) x + 1 = 0

For a quadratic equation ax² + bx + c = 0, the discriminant is:

D = b² − 4ac

For real and equal roots, we set:

D = 0 

Identify the coefficients

From the equation:

a = p + 1

b = −2(p − 1)

c = 1

D = [−2 (p − 1)]² − 4(p + 1) (1)

[−2(p − 1)]²

= 4(p − 1)²

= 4(p² − 2p + 1)

4(p + 1) (1)

= 4(p + 1)

D = 4(p² − 2p + 1) − 4(p + 1)

= 4p² − 12p

4p² − 12p = 0

Divide both sides by 4:

p² − 3p = 0

⇒ p(p − 3) = 0

p = 0 or p = 3​