Question

Let $x^3+a{x}^2+bx+c=0$ has roots $\alpha ,\beta ,\gamma$. If $\alpha +2=\frac{1}{{\alpha }^2},\beta +2=\frac{1}{{\beta }^2}$ and $\gamma +2=\frac{1}{{\gamma }^2}$, then the value of $3a+2b+c$ is

Answer 1

As the given relation is
$\alpha +2=\frac{1}{{\alpha }^2}\Rightarrow {\alpha }^3+2{\alpha }^2-1=0\cdots \left(1\right)$
Similarly,
${\beta }^3+2{\beta }^2-1=0\cdots \left(2\right)$
${\gamma }^3+2{\gamma }^2-1=0\cdots \left(3\right)$
From equations $\left(1\right),\left(2\right)$ and $\left(3\right),$ we have
$x^3+2x^2-1=0\cdots \left(4\right)$
And $\alpha ,\beta ,\gamma$ are the roots of equation $\left(4\right).$

Now, $x^3+a{x}^2+bx+c=0$ has roots $\alpha ,\beta ,\gamma$, so on comparing this equation with eq. $\left(4\right),$ we get
$a=2,b=0,c=-1$
$\therefore 3a+2b+c=5$