Question

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 is

Answer 1

$x^2-(a-2)x-a-1=0\Rightarrow \alpha +\beta =a-2\alpha \beta =-(a+1){\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =a^2-2a+6=(a-1{)}^2+5\Rightarrow a=1$