Question

Solve $18x^3+81x^2+121x+60=0$ given that the root is equal to half the sum and the remaining roots?

Answer 1

$18x^3+81x^2+121x+60=0$

One root is the half of sum of remaining roots. So roots are in AP.

Let roots be $\alpha -\beta ,\alpha ,\alpha +\beta$

Sum of roots $=-\frac{81}{18}=3\alpha$

$\alpha =-\frac{3}{2}$

So one root is $x=-\frac{3}{2}$

Dividing $18x^3+81x^2+121x+60=0$ by $x+\frac{3}{2}$ we get $18x^2+54x+40$

Hence other roots are given by $18x^2+54x+40=0$

$⟹9x^2+27x+20=0$

$x=\frac{-27\pm \sqrt{{27}^2-(4\ast 9\ast 20)}}{2\ast 9}=\frac{-27\pm 3}{18}=-\frac{4}{3},-\frac{5}{3}$

The roots are $-\frac{3}{2},-\frac{4}{3},-\frac{5}{3}$