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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.
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For real roots, $D \geq 0$. $[-2(p+1)]^2 - 4p^2 \geq 0 \Rightarrow p \geq -\frac{1}{2}$ ($\frac{1}{2} + \frac{1}{2} + 1$ marks). $\therefore$ smallest value of $p = -\frac{1}{2}$ ($\frac{1}{2}$ mark). At $p = -\frac{1}{2}$ given equation becomes $x^2 - 2(-\frac{1}{2} + 1)x + (-\frac{1}{2})^2 = 0$ ($\frac{1}{2}$ mark). $x^2 - x + \frac{1}{4} = 0$ or $4x^2 - 4x + 1 = 0$ (1 mark). $(2x-1)(2x-1) = 0$. $\therefore$ roots are $\frac{1}{2}, \frac{1}{2}$ ($\frac{1}{2} + \frac{1}{2}$ marks).