Question

If $\alpha ,\beta$ are the roots of $x^2-(a-2)x-(a+1)=0,$ where ${}^{\prime }{a}^{\prime }$ is a parameter, then the minimum value of ${\alpha }^2+{\beta }^2$ is equal to

Answer 1

$\alpha ,\beta$ are the roots of $x^2-(a-2)x-(a+1)=0$
$\Rightarrow \alpha +\beta =a-2,\alpha \cdot \beta =-(a+1)$
Let $f\left(a\right)={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \cdot \beta$
$=(a-2{)}^2+2(a+1)$
$\Rightarrow f\left(a\right)=a^2-2a+6$
$\Rightarrow f^{\prime }\left(a\right)=2a-2$ and $f^{\prime \prime }\left(a\right)=2>0$
$\therefore f\left(a\right)$ has minimum at $f^{\prime }\left(a\right)=0$
$\Rightarrow a=1$
$f_{min}\left(a\right)=(1{)}^2-2(1)+6=5$