Question

The value of p for which the sum of the squares of the roots of the equation $x^2-(p-2)x-p-1=0$ assume the least value is:

• A. -1
• B. 1
• C. 2

Answer 1

The correct option is B 1

Let $\alpha ,\beta$ be the roots of the given equation, then, $\alpha +\beta =p-2and\alpha \beta =-(p+1).$
Now, ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(p-2{)}^2+2(p+1)$
$=p^2-2p+6=(p-1{)}^2+5$
Clearly ${\alpha }^2+{\beta }^2\ge 5$. So minimum value of ${\alpha }^2+{\beta }^2$ is 5, which attains at p = 1.