If the sum of two roots $\alpha, \beta$ of the equation $x^4-x^3-8 x^2+$ $2 x+12=0$ is zero and $\gamma,…
If the sum of two roots $\alpha, \beta$ of the equation $x^4-x^3-8 x^2+$ $2 x+12=0$ is zero and $\gamma, \delta(\gamma\gt\delta)$ are its other roots, then $3 \gamma+2 \delta=$
$0$
$1$
$3$
$5$
Solution
Given $\alpha, \beta, \gamma$ and $\delta$ be the roots of $x^4-x^3-8 x^2+2 x+12=0$. So, such that $\alpha+\beta=0$
$\begin{aligned} & x^4-x^3-8 x^2+2 x+12=\left(x^2+a\right)\left(x^2-x+b\right) \\ & =x^4-x^3+(a+b) x^2-a x+a b\end{aligned}$
After equating co-efficient of $x$ and constant term, we get
$\begin{aligned} & -a=2 \Rightarrow a=-2 \text { and } a b=12 \Rightarrow-2 b=12 \Rightarrow b=-6 \\ & \text { So, } x^4-x^3-8 x^2+2 x+12=\left(x^2-2\right)\left(x^2-x-6\right) \\ & =\left(x^2-2\right)(x-3)(x+2)\end{aligned}$
hence, $\gamma=3, \delta=-2$ $(\because \gamma\gt\delta)$
Now, $3 \gamma+2 \delta=9-4=5$