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If one root of the equation $x^3-6 x^2+3 x+10=0$ is the average of the other two, then the sum of fourth powers of the roots of the equation is
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The correct answer is:
$642$
Let $\alpha, \beta$ and $\gamma$ are the roots of $x^3-6 x^2+3 x+10=0$
Also, let $\beta=\frac{\alpha+\gamma}{2} \Rightarrow 2 \beta=\alpha+\gamma$
$\therefore \alpha+2+\gamma=6 \Rightarrow \alpha+\gamma=4$ ...(i)
Also, $\alpha \beta \gamma=-10 \Rightarrow \alpha \gamma=-5$
Now, $\alpha-\gamma=\sqrt{(\alpha+\gamma)^2-4 \alpha \gamma}=\sqrt{16+20}= \pm 6$
$\begin{aligned} & \text { Equation (i) + (ii) } \\ & 2 \alpha=10 \Rightarrow \alpha=5 \\ & \text { from (i) } \gamma=4-5=-1 \\ & \text { Now, } \alpha^4+\beta^4+\gamma^4=5^4+2^4+(-1)^4=625+16+1 \\ & =642\end{aligned}$
Also, let $\beta=\frac{\alpha+\gamma}{2} \Rightarrow 2 \beta=\alpha+\gamma$
$\therefore \alpha+2+\gamma=6 \Rightarrow \alpha+\gamma=4$ ...(i)
Also, $\alpha \beta \gamma=-10 \Rightarrow \alpha \gamma=-5$
Now, $\alpha-\gamma=\sqrt{(\alpha+\gamma)^2-4 \alpha \gamma}=\sqrt{16+20}= \pm 6$
$\begin{aligned} & \text { Equation (i) + (ii) } \\ & 2 \alpha=10 \Rightarrow \alpha=5 \\ & \text { from (i) } \gamma=4-5=-1 \\ & \text { Now, } \alpha^4+\beta^4+\gamma^4=5^4+2^4+(-1)^4=625+16+1 \\ & =642\end{aligned}$
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