Question

Solve the equations:
$3x^3-26x^2+52x-24=0$, the roots being in geometrical progression.

Answer 1

${3x}^3{-26x}^2+52x-24=0$

Let the roots be $\frac{a}{r},a,ar$

$\sum xyz=\frac{a}{r}\left(a\right)\left(ar\right)=-\frac{-24}{3}=8\Rightarrow a^3=8\Rightarrow a=2\sum x=\frac{a}{r}+a+ar=-\frac{-26}{3}a\left(\frac{1+r+r^2}{r}\right)=\frac{26}{3}2\left(\frac{1+r+r^2}{r}\right)=\frac{26}{3}\Rightarrow 3+3r+3r^2=13r\Rightarrow 3r^2-10r+3=0\Rightarrow {3r}^2-9r-r+3=0\Rightarrow 3r(r-3)-1(r-3)=0r=3,\frac{1}{3}$

So the roots of the equation are $\frac{2}{3},2,6$