Question

Solve the equations:
$54x^3-39x^2-26x+16=0$, the roots being in geometrical progression.

Answer 1

$54x^3-39x^2-26x+16=0$
Let the roots of the equation be $\frac{a}{r},a,ar$

$S_3=\frac{a}{r}\left(a\right)\left(ar\right)=-\frac{16}{54}\Rightarrow a^3=-\frac{8}{27}\Rightarrow a=-\frac{2}{3}{S}_1=a(1+r+\frac{1}{r})=-\frac{-39}{54}\Rightarrow -\frac{2}{3}\left(\frac{r+r^2+1}{r}\right)=\frac{13}{18}\Rightarrow 12r^2+25r+12=0\Rightarrow 12r^2+16r+9r+12=0\Rightarrow 4r(3r+4)+3(3r+4)=0\Rightarrow r=-\frac{4}{3},-\frac{3}{4}$

So the roots of the equation are $\frac{1}{2},\frac{-2}{3},\frac{8}{9}$