Question

If $\alpha$ and $\beta$ are roots of the equation $x^{2}+p x+\frac{3 p}{4}=0$, suchThat $|\alpha-\beta|=\sqrt{10}$, then $p$ belongs to the se :

• A. {2, - 5}
• B. {- 3, 2}
• C. {- 2, 5}
• D. {3, - 5}

Answer 1

Answer: (C)

Given $\alpha, \beta$ are roots of equation
$x^{2}+p x+3 p / 4=0$
$(\alpha+\beta)=-p ; \alpha \cdot \beta=3 p / 4$
$\Rightarrow (\alpha-\beta)=\sqrt{10}$
$\Rightarrow (\alpha-\beta)^{2}=10$
$\Rightarrow (\alpha+\beta)^{2}-4 \alpha \beta=10$
$p^{2}-4(3 p / 4)=10$
$p^{2}-3 p-10=0$
$p^{2}-5 p+2 p-10=0$
$(p-5)(p+2)=0$
$p=-2,5$
$P \in\{-2,5\}$