Question

Find the value of a for which the sum of the squares of the roots of the equation $x^2-(a-2)x-a-1=0$ assumes the least value.

• A. $a=1$
• B. $a=-1$
• C. $a=0$
• D. $a=2$

Answer 1

Answer: (A)

$a=1$

Given equation is
$x^2-(a-2)x-a-1=0$
$\alpha +\beta =a-2$
$\alpha \beta =-a-1$
Now, $S={\alpha }^2+{\beta }^2$
$\Rightarrow S=(\alpha +\beta {)}^2-2\alpha \beta$
$\Rightarrow S=(a-2{)}^2+2a+2$
$\Rightarrow S=a^2-2a+6$
Since, S assumes the least value
$\Rightarrow \frac{dS}{da}=0$
$\Rightarrow 2a-2=0$
$\Rightarrow a=1$
Also, $\frac{d^{2}S}{d^2}=2>0$ . Hence $a=1$