Question

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

x² − 2(p + 1) x + p² = 0

Answer 1

Given:

x² − 2(p + 1) x + p² = 0

For any quadratic equation of the form:

ax² + bx + c = 0

D = b² − 4ac

For real and equal roots, the condition is:

D = 0

Identify coefficients

From the equation:

a = 1

b = −2(p+1)

c = p²

Write and simplify the discriminant

D = [−2(p + 1)]² − 4(1) (p²)

= 4(p + 1)² − 4p²

= 4(p² + 2p + 1) − 4p²

= 4p² + 8p + 4 − 4p²

= 8p + 4

= 4(p² + 2p + 1) − 4p²

= 4p² + 8p + 4 − 4p²

= 8p + 4

8p + 4 = 0

⇒ 8p = −4

p = `-1/2`

This is the value of p for which the equation has real and equal roots.