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Q.
For what value of $a$ the sum of squares of the roots of the equation $x^2-(a-2) x-(a+1)=0$ is least?
Complex Numbers and Quadratic Equations
Solution:
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., $a=1$.