Question

Fnd 3a if $a,-1,-\frac{1}{a}$ are the zeroes of the cubic polynomial $p\left(x\right)=3x^3-5x^2-11x-3$. a>0

Answer 1

Given,

$a,-1,-\frac{1}{a}$ are the zeroes of the cubic polynomial $p\left(x\right)=3x^3-5x^2-11x-3$.

Then we've from the relation between the roots and co-efficient we get,

$a-1-\frac{1}{a}=\frac{5}{3}$

or, $a-\frac{1}{a}=\frac{8}{3}$

or, $3(a^2-1)=8a$

or, $3a^2-8a-3=0$

or, $3a^2-9a+a-3=0$

or, $(3a+1)(a-3)=0$

or, $a=-\frac{1}{3}$ or, $a=3$.

$3a=-3$ or, $3a=9$.