Question

If $\alpha$, $\beta$, $\gamma$ are zeroes of cubic polynomial $x^3+p{x}^2+q{x}^2+2$ such that $\alpha \beta +1=0$.
Find the value of $2p+q+5$.

Answer 1

$\alpha ,\beta ,\gamma$ zeroes of polynomial $x^3+p{x}^2+q{x}^2+2$ i.e, on keeping $x=\alpha or\beta or\gamma ,$ we will get $x^3+px+qx+2=0$

Also, $\alpha \beta +1=0$

We know that sum of roots of cubic polynomial $=\frac{-p}{1}$

$\Rightarrow \alpha +\beta +\gamma =-p$ $\&$ $\alpha \beta +\beta \gamma +\gamma \alpha =q$

$\Rightarrow \alpha \beta \gamma =-2$

Since $\alpha \beta \gamma =-2$

$\&$ $\alpha \beta +1=0\Rightarrow \alpha \beta =-1$

$\Rightarrow (-1)\gamma =-2$

$\Rightarrow \gamma =2$

$\therefore \alpha +\beta +2=-p$ $\Rightarrow \alpha +\beta =-p-2\to \left(1\right)$

$\alpha \beta +\beta \gamma +\gamma \alpha =q\Rightarrow -1+2\beta +2\alpha =q$

$\Rightarrow (\alpha +\beta )=\frac{(q+1)}{2}\to \left(2\right)$

Equating $\left(1\right)\&\left(2\right)$

$\Rightarrow -p-2=\frac{q+1}{2}$

$\Rightarrow -2p-4=q+1$

$\Rightarrow 2p+q+5=0$

Hence, the answer is $2p+q+5=0.$