Question

Find the smallest value of p for which the quadratic equation x² − 2(p + 1)x + p² = 0 has real roots. Hence, find the roots of the equation so obtained.

Answer 1

x² − 2(p + 1)x + p² = 0

a = 1, b = −2(p + 1), c = p²

D = b² − 4ac

= [−2(p + 1)]² − 4 (1) (p²)

= 4(p + 1)² − 4p²

= 4p² + 4 + 8p − 4p²

If roots are real.

D ≥ 0

4 + 8p ≥ 0

8p ≥ −4

`p ≥ -4/8`

`p ≥ -1/2`

∴ Smallest value of p is `-1/2`.

Quadratic equation, x² − 2(p + 1)x + p² = 0

⇒ x² − 2px + p² − 2x = 0

⇒ (x − p)² − 2x = 0

Put p = `-1/2`

⇒ `(x + 1/2)^2 - 2x = 0`

⇒ `x^2 + x + 1/4 - 2x = 0`

⇒ `x^2 - x + 1/4 = 0`

⇒ 4x² − 4x + 1 = 0

⇒ (2x − 1)² = 0

⇒ (2x − 1) (2x − 1) = 0

x = `1/2` and x = `1/2`