Question

Find the value of $a$ such that the sum of the squares of the roots of the equation $x^{2}-(a-2) x-(a+1)=0$ is least.

• A. 4
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (C)

Let $\alpha, \beta$ be the roots of the equation.
$\therefore \alpha+\beta=a-2$ and $\alpha \beta=-(a+1)$
Now, $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$=(a-2)^{2}+2(a+1)=(a-1)^{2}+5$
$\therefore \alpha^{2}+\beta^{2}$ will be minimum if $(a-1)^{2}=0$,
i.e. $a=1$.