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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4 x^2-9 x-36$ $=0$ and $\alpha < \beta < \gamma$ then $\alpha+2 \beta+3 \gamma=$
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$-1$
$x^3+4 x^2-9 x-36=0$
$\Rightarrow \quad x^2(x+4)-9(x+4)=0$
$\begin{aligned} & \Rightarrow \quad(x+4)(x+3)(x-3)=0 \\ & \therefore \quad x=-4,-3,3 \Rightarrow \alpha=-4, \beta=-3, \gamma=3 \\ & \Rightarrow \quad \alpha+2 \beta+3 \gamma=-4+2(-3)+3(3)=-1 .\end{aligned}$
$\Rightarrow \quad x^2(x+4)-9(x+4)=0$
$\begin{aligned} & \Rightarrow \quad(x+4)(x+3)(x-3)=0 \\ & \therefore \quad x=-4,-3,3 \Rightarrow \alpha=-4, \beta=-3, \gamma=3 \\ & \Rightarrow \quad \alpha+2 \beta+3 \gamma=-4+2(-3)+3(3)=-1 .\end{aligned}$
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