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If $\alpha, \beta, \gamma$ are the roots of $x^3-3 x^2-4 x+12=0$, then $\sum(\alpha+\beta)^2$ is equal to
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The correct answer is:
26
Given, $x^3-3 x^2-4 x+12=0$
$\begin{aligned}
& \Rightarrow \quad(x-2)\left(x^2-x-6\right)=0 \\
& \Rightarrow \quad(x-2)(x+2)(x-3)=0 \\
& \Rightarrow \quad x=-2,2,3 \\
& \text { So, } \quad \alpha=-2, \beta=2 \text { and } \gamma=3 \\
& \therefore \quad \Sigma(\alpha+\beta)^2=(\alpha+\beta)^2+(\beta+\gamma)^2+(\gamma+\alpha)^2 \\
& =0+(5)^2+(1)^2=25+1=26 \\
&
\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad(x-2)\left(x^2-x-6\right)=0 \\
& \Rightarrow \quad(x-2)(x+2)(x-3)=0 \\
& \Rightarrow \quad x=-2,2,3 \\
& \text { So, } \quad \alpha=-2, \beta=2 \text { and } \gamma=3 \\
& \therefore \quad \Sigma(\alpha+\beta)^2=(\alpha+\beta)^2+(\beta+\gamma)^2+(\gamma+\alpha)^2 \\
& =0+(5)^2+(1)^2=25+1=26 \\
&
\end{aligned}$
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