Question

Solve $3x^3-26x^2+52x-24=0$, given that the roots are in geometric progression

Answer 1

Let $\frac{a}{r},a$ and $ar$ are the roots of $3x^3-26x^2+52x-24=0$

Thus products of roots $=\frac{24}{3}=a^3\Rightarrow a=2$

And sum of roots $=\frac{26}{3}=\frac{2}{r}+2+2r$

$\Rightarrow \frac{1}{r}+r=\frac{10}{3}$

$\Rightarrow 3r^2-10r+3=0$

$\Rightarrow (3r-1)(r-3)=0$

$\Rightarrow r=\frac{1}{3},3$

Hence the roots are $\frac{2}{3},2,6$