If one of the roots of $18x^3+81x^2+121x+60=0$ is equal to half the sum of the other two,  then one of its roots is

The given equation is $18x^3+81x^2+121x+60=0$ 

$\alpha ,\beta ,\gamma$ are the roots of given equation

By the given condition, one of the roots of equation is equal to half the sum of the other two roots

i.e., $\alpha =\frac{\beta +\gamma }{2}\Rightarrow \beta +\gamma =2\alpha$

$\therefore s_1=\alpha +\beta +\gamma =\frac{-81}{18}$

$\Rightarrow \text{3}\alpha =\frac{-9}{2}$

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

$\therefore$  one of the root is $-\frac{3}{2}$