Question

Solve the equations:
$18x^3+81x^2+121x+60=0$, one root being half the sum of the other two.

Answer 1

$18x^4+81x^3+121x+60=0$
Let the roots of the equation be $a,b,\frac{a+b}{2}$

$S_1=a+b+\frac{a+b}{2}=-\frac{81}{18}\frac{3a+3b}{2}=-\frac{9}{2}\Rightarrow a+b=-\mathrm{3.....}\left(i\right)S_4=ab\left(\frac{a+b}{2}\right)=-\frac{60}{18}ab\left(\frac{-3}{2}\right)=-\frac{60}{18}ab=\frac{20}{9}.....\left(ii\right)$

substituting $b$ from $\left(i\right)$

$a(-3-a)=\frac{20}{9}{a}^2+3a+\frac{20}{9}=09a^2+27a+20=09a^2+12a+15a+20=03a(3a+4)+5(3a+4)=0(3a+4)(3a+5)=0\Rightarrow a=-\frac{4}{3},-\frac{5}{3}$

substituting $a$ in $\left(ii\right)$

$\Rightarrow b=-\frac{5}{3},-\frac{4}{3}$

So the roots of the equation are $-\frac{5}{3},-\frac{4}{3},-\frac{3}{2}$