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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-4 x^2-9 x-36$ $=0$ such that $\alpha+\beta=0$, then $\alpha^2+2 \beta^2+3 \gamma^2=$
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The correct answer is:
75
Given, equation $x^3+4 y^2-9 x-36=0$ having roots $\alpha, \beta$ and $\gamma$
$$
\begin{aligned}
& \because \alpha+\beta=0 \Rightarrow \gamma=-4 \\
& \Rightarrow \alpha=-\beta \\
& \text { Now } \alpha^2+2 \beta^2+3 \gamma^2=3\left(\beta^2+\gamma^2\right) \\
& =3\left(3^2+(-4)^2\right) \quad \text{[from equation (ii) $\beta=3$ ]} \\
& =75
\end{aligned}
$$
$$
\begin{aligned}
& \because \alpha+\beta=0 \Rightarrow \gamma=-4 \\
& \Rightarrow \alpha=-\beta \\
& \text { Now } \alpha^2+2 \beta^2+3 \gamma^2=3\left(\beta^2+\gamma^2\right) \\
& =3\left(3^2+(-4)^2\right) \quad \text{[from equation (ii) $\beta=3$ ]} \\
& =75
\end{aligned}
$$
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