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

Given:
Quadratic equation: \( x^2 - 2(p + 1)x + p^2 = 0 \)

We are to find the smallest integer value of \( p \) for which the equation has real roots and also find the roots.

Step 1: Identify coefficients
The standard quadratic equation is \( ax^2 + bx + c = 0 \)
Comparing with the given equation:
- \( a = 1 \)
- \( b = -2(p + 1) \)
- \( c = p^2 \)

Step 2: Use the discriminant condition
A quadratic equation has real roots if the discriminant \( D \geq 0 \), where:
\[ D = b^2 - 4ac \]
Substitute values of \( a, b, c \):
\[ D = [ -2(p + 1) ]^2 - 4 \cdot 1 \cdot p^2 = 4(p + 1)^2 - 4p^2 = 4[(p + 1)^2 - p^2] \]
Step 3: Simplify the expression inside the bracket
\[ (p + 1)^2 = p^2 + 2p + 1 \Rightarrow (p + 1)^2 - p^2 = 2p + 1 \Rightarrow D = 4(2p + 1) \]
Step 4: Apply condition for real roots
Since \( D \geq 0 \), we get:
\[ 4(2p + 1) \geq 0 \Rightarrow 2p + 1 \geq 0 \Rightarrow p \geq -\frac{1}{2} \]
Step 5: Choose the smallest integer value of \( p \)
Since \( p \geq -\dfrac{1}{2} \), the smallest integer value of \( p \) is:
\[ \boxed{p = 0} \]
Step 6: Find the roots by substituting \( p = 0 \)
Substitute into the original equation:
\[ x^2 - 2(p + 1)x + p^2 = x^2 - 2(0 + 1)x + 0^2 = x^2 - 2x = 0 \Rightarrow x(x - 2) = 0 \Rightarrow x = 0 \text{ or } x = 2 \]
Final Answer:
Smallest integer value of \( p = \boxed{0} \)
The roots of the equation are: \( \boxed{x = 0 \text{ and } x = 2} \)