If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-3 x^2+x+5=0$, then $y=\Sigma \alpha^2+\alpha \beta \gamma$ satisfies the equation

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Aicharya · Accepted Answer

Given, $\alpha, \beta$ and $\gamma$ are roots of $x^3-3 x^2+x+5=0$ $	herefore \quad \alpha+\beta+\gamma=3$ $\alpha \beta+\beta \gamma+\gamma \alpha=1$ $\alpha \beta \gamma=-5$ Now, $y=\Sigma \alpha^2+\alpha \beta \gamma$ $\begin{aligned} & y=\left(\alpha^2+\beta^2+\gamma^2ight)+\alpha \beta \gamma \ & y=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)+\alpha \beta \gamma \ & y=(3)^2-2(1)-5=9-7=2\end{aligned}$ $	herefore \quad y=2$ $y=2$ only satisfied the equation $y^3-y^2-y-2=0$

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-3 x^2+x+5=0$, then $y=\Sigma…

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