Question

If the absolute difference between two roots of the equation $x^2+px+3=0$ is $\sqrt{p}$, then $p$ equals

• A. $-3,4$
• B. $4$
• C. $-3$
• D. None of these

Answer 1

Answer: (B)

$4$

Given quadratic equation
$x^2+px+3=0$
Let $\alpha ,\beta$ be the roots of the equation
$\alpha +\beta =-p$
$\alpha \beta =3$
Also, $(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$
$\Rightarrow (\alpha -\beta {)}^2=p^2-12$
Given, $|\alpha -\beta |=\sqrt{p}$
$(\alpha -\beta {)}^2=p$
$p^2-12=p$
$p^2-p-12=0$
$p=4,-3$
But $p$ cannot be negative.
So, $p=4$