If and are roots of x^2+px+3p4=0 such that |-|=√10, then p belongs to the set:

Question

If $\alpha$ and $\beta$ are roots of $x^2+px+\dfrac{3p}4=0$ such that $|\alpha-\beta|=\sqrt{10}$, then $p$ belongs to the set:

Accepted Answer

$\alpha+\beta=-p$, $\alpha\beta=\dfrac{3p}4$.

$(\alpha-\beta)^2=p^2-3p=10 \implies p^2-3p-10=0 \implies p=-2,5.$

Solutions · 1