Question

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

Answer 1

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