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If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4-8 x^3+11 x^2+32 x-60=0$ and $\alpha < \beta < \gamma < \delta$, then $4 \alpha+3 \beta+2 \gamma+\delta=$
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The correct answer is:
$9$
We have,
$\begin{aligned} x^4-8 x^3+11 x^2+32 x-60 & =0 \\ \Rightarrow \quad(x+2)(x-2)(x-3)(x-5) & =0\end{aligned}$
$\therefore \quad \alpha=-2, \beta=2, \gamma=3, \delta=5$
$\begin{aligned} \therefore \quad 4 \alpha+3 \beta+2 \gamma+\delta & =4(-2)+3(2)+2(3)+5 \\ & =-8+6+6+5=9\end{aligned}$
$\begin{aligned} x^4-8 x^3+11 x^2+32 x-60 & =0 \\ \Rightarrow \quad(x+2)(x-2)(x-3)(x-5) & =0\end{aligned}$
$\therefore \quad \alpha=-2, \beta=2, \gamma=3, \delta=5$
$\begin{aligned} \therefore \quad 4 \alpha+3 \beta+2 \gamma+\delta & =4(-2)+3(2)+2(3)+5 \\ & =-8+6+6+5=9\end{aligned}$
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