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

Hint: At first take roots as \[\alpha ,\beta ,\gamma \] then according to the problem write \[\beta \] as \[\dfrac{1}{2}\left( \alpha +\gamma \right)\], then substitute in relation that, \[\alpha +\beta +\gamma =\dfrac{-b}{a}\] and find \[\beta \], after getting one of the roots find others by dividing the main equation by \[x-\beta \] and further factorize it to get results.

Complete step-by-step answer:
In the question we are given an equation \[18{{x}^{3}}+81{{x}^{2}}+121x+60=0\] and we have to find the values of x given situation that the root is equal to the half the sum of remaining roots.
Now before proceeding we will write about the relation of roots with the equation.
So, if the equation is in form of \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] and roots be \[\alpha ,\beta ,\gamma \] then we can say that,
\[\alpha +\beta +\gamma =\dfrac{-b}{a}\]
\[\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}\]
And \[\alpha \beta \gamma =\dfrac{-d}{a}\]
Now, here in the question the equation given is \[18{{x}^{3}}+81{{x}^{2}}+121x+60=0\] and let’s suppose that \[\alpha ,\beta ,\gamma \] are roots then we can say that,
\[\begin{align}
  & \alpha +\beta +\gamma =\dfrac{-81}{18}=\dfrac{-9}{2} \\
 & \alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{121}{18} \\
\end{align}\]
And \[\alpha \beta \gamma =\dfrac{60}{18}=\dfrac{10}{3}\]
Now, furthermore one more condition is given that one root is half of the sum of remaining roots. So, we can suppose that,
\[\beta =\dfrac{\alpha +\gamma }{2}\]
So, now we will use the relation, \[\beta =\dfrac{\alpha +\gamma }{2}\] or \[\alpha +\gamma =2\beta \] to proceed in \[{{1}^{st}}\] relation which is,
\[\alpha +\beta +\gamma =\dfrac{-9}{2}\]
Or, \[\left( \alpha +\gamma \right)+\beta =\dfrac{-9}{2}\]
Or, \[2\beta +\beta =\dfrac{-9}{2}\]
So, \[3\beta =\dfrac{-9}{2}\]. Hence, the value of \[\beta \] is \[\dfrac{-3}{2}\].
So, now we know that one of the roots is \[\dfrac{-3}{2}\] hence then we can say that the equation is divisible by \[\left( x-\left( \dfrac{-3}{2} \right) \right)\] or \[\left( x+\dfrac{3}{2} \right)\].
So now we can write, \[18{{x}^{3}}+81{{x}^{2}}+121x+60=0\] as \[18{{x}^{3}}+27{{x}^{2}}+54{{x}^{2}}+81x+40x+60=0\].
Hence, on factoring we get,
\[18{{x}^{2}}\left( x+\dfrac{3}{2} \right)+54x\left( x+\dfrac{3}{2} \right)+40\left( x+\dfrac{3}{2} \right)=0\]
So we can write it as,
\[\left( x+\dfrac{3}{2} \right)\left( 18{{x}^{2}}+54x+40 \right)=0\]
Or, \[\left( x+\dfrac{3}{2} \right)\left( 18{{x}^{2}}+24x+30x+40 \right)=0\]
Hence on factoring we can write it as,
\[\left( x+\dfrac{3}{2} \right)\left( 18x\left( x+\dfrac{4}{3} \right)+30\left( x+\dfrac{4}{3} \right) \right)=0\]
Or, \[\left( x+\dfrac{3}{2} \right)\left( 18x+30 \right)\left( x+\dfrac{4}{3} \right)=0\]
Hence values of x for which equation satisfies is \[\dfrac{-3}{2},\dfrac{-4}{3},\dfrac{-30}{18}\] or \[\dfrac{-5}{3}\].
Hence, the roots are: \[\dfrac{-3}{2},\dfrac{-4}{3}\] and \[\dfrac{-5}{3}\].

Note: Students can also do the problem by taking roots as \[\alpha -\beta ,\alpha ,\alpha +\beta \] according to the given condition in the question. And after finding the first they can directly find other two roots by using formula \[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] where a, b are coefficients of \[{{x}^{2}},x\] and c is constant in quadratic equation \[a{{x}^{2}}+bx+c=0\].