Question

If the sum of the roots of an equation $x^2+px+1=0(p>0)$ is twice the difference between them, then $p=$ ____.

Answer 1

Sum $=\alpha +\beta =-p$

Difference $=\alpha -\beta =\sqrt{{(\alpha +\beta )}^2-4\alpha \beta }$

$=\sqrt{p^2-4}$

According to the question,

$\Rightarrow 2\sqrt{p^2-4}=-p$

$\Rightarrow 4(p^2-4)=p^2$

$\Rightarrow {4p}^2-p^2=16\Rightarrow {3p}^2=16$

$\Rightarrow p=\sqrt{\frac{16}{3}}=\pm \frac{4}{\sqrt{3}}$

Hence, this is the answer.