Question

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

Hint:

For real roots, ensure discriminant $D \geq 0$. Simplify step-by-step before solving.

Answer 1

Problem:
Solve the quadratic equation \( x^2 - 2(p + 1)x + p^2 = 0 \) for real roots and find the smallest integer value of \( p \).

Solution:

1. Identify Coefficients:
Given \( ax^2 + bx + c = 0 \), then:
- \( a = 1 \)
- \( b = -2(p + 1) \)
- \( c = p^2 \)

2. Discriminant Condition for Real Roots:
Real roots exist if the discriminant \( D \geq 0 \), where:
\[D = b^2 - 4ac\]
Substitute values:
\[D = [ -2(p + 1) ]^2 - 4 \cdot 1 \cdot p^2 = 4(p + 1)^2 - 4p^2 = 4[(p + 1)^2 - p^2]\]
3. Simplify the Discriminant:
\[(p + 1)^2 = p^2 + 2p + 1 \Rightarrow (p + 1)^2 - p^2 = 2p + 1 \Rightarrow D = 4(2p + 1)\]
4. Apply the Condition \( D \geq 0 \):
\[4(2p + 1) \geq 0 \Rightarrow 2p + 1 \geq 0 \Rightarrow p \geq -\frac{1}{2}\]
5. Find the Smallest Integer Value of \( p \):
Since \( p \geq -\dfrac{1}{2} \), the smallest integer value of \( p \) is:
\[\boxed{p = 0}\]
6. Find the Roots for \( p = 0 \):
Substitute \( p = 0 \) into the original equation:
\[x^2 - 2(p + 1)x + p^2 = x^2 - 2(0 + 1)x + 0^2 = x^2 - 2x = 0 \Rightarrow x(x - 2) = 0 \Rightarrow x = 0 \text{ or } x = 2\]
Answer:
Smallest integer value of \( p = \boxed{0} \)
Roots: \( \boxed{x = 0 \text{ and } x = 2} \)