Question

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2+(p-3) x-2 p=3(p \in R)$, then the minimum value of $\left(\alpha^2+\beta^2+\alpha \beta\right)$, is

• A. 2
• B. 4
• C. 8
• D. 12

Answer 1

Answer: (C)

$\alpha+\beta=-(p-3)$
$\alpha \beta=-2 p-3$
$\because \alpha^2+\beta^2+\alpha \beta= (\alpha+\beta)^2-\alpha \beta $
$=(p-3)^2+2 p +3$
$ = p ^2-4 p +12 $
$ =( p -2)^2+8$
$\therefore $ minimum value of $\alpha^2+\beta^2+\alpha \beta$ will be 8 .