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

Given:

x² − 2(p + 1)x + p² = 0

Step 1: Condition for Real Roots

For a quadratic equation ax² = bx + C = 0 to have real roots, its discriminant must be non-negative:

Δ = b² − 4ac ≥ 0

Comparing with the given equation:

• a = 1,
• b = −2(p + 1)
• c = p²

The discriminant is:

Δ = [−2(p + 1)]² −4(1)(p²)

= 4(p + 1)² − 4p²

= 4(p² + 2p + 1) − 4p²

= 8p + 4 ≥ 0

Step 2: Solve for p

8p + 4 ≥ 0

8p ≥ −4

p ≥ `-1/2`

Since we need the smallest integer value of p, we take

p = 0

Step 3: Find the Roots for p = 0

Substituting p = 0 in the quadratic equation:

x² − 2(0 + 1)x + 0² = 0

x² − 2x = 0

Factorising:

x(x − 2) = 0

x = 0 or x = 2

Smallest value of ppp for real roots: p = 0.

Roots of the equation for p = 0: x = 0 and x = 2.