Question

The value of $a$ for which the sum of the squares of the root of the equation
$x^2-(a-2)x-a-1=0$ assumes the least value, is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

Let the roots of the equation are $\alpha ,\beta$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha .\beta =(a-2{)}^2+2(a+1)=(a-1{)}^2+5$

So, ${\alpha }^2+{\beta }^2\ge 5$, and having the minimum value $5$ at $a=1$.