Question

The equation $x^3-6x^2+px-8=0$ has only positive real roots. The value of $p$ is

Answer 1

Answer: 12

Let $r,s,t$ be positive roots
$r+s+t=6;rst=8$
$\Rightarrow A.M.=\frac{r+s+t}{3}=2$
and $G.M.=\sqrt[3]{rst}=2;$
$\Rightarrow AM=GM$
Hence $r=s=t=2$
$\Rightarrow rs+st+tr=p$;
$\therefore p=12$