Question

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

• A. $\left\{2, -5\right\}$
• B. $\left\{-3, 2\right\}$
• C. $\left\{-2, 5\right\}$
• D. $\left\{3, -5\right\}$

Answer 1

Answer: (C)

Given quadratic eqn. is
$x^{2}+px+\frac{3p}{4}=0$
So, $\alpha+\beta=-p, \alpha\beta=\frac{3p}{4}$
Now, given $\left|\alpha -\beta \right|=\sqrt{10}$
$\Rightarrow \alpha -\beta=\pm \sqrt{10}$
$\Rightarrow \left(\alpha -\beta \right) ^{2}=10 \Rightarrow \alpha^{2} +\beta^{2}-2\alpha\beta =10$
$\Rightarrow \left(\alpha +\beta \right)^{2}-4\alpha\beta =10$
$\Rightarrow p^{2}-4\times\frac{3p}{4}=10 \Rightarrow p^{2}-3p-10=0$
$\Rightarrow p=-2, 5 \Rightarrow p\, \in\,\left\{-2, 5\right\}$