The sum of two roots of the equation x^4-x^3- | vedclass

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(B) Let the roots be $\alpha, \beta, \gamma, \delta$ such that $\alpha+\beta=0$,which implies $\beta=-\alpha$. Since $\alpha$ and $-\alpha$ are roots,

Vedclass · Accepted Answer

$369$

Vedclass · Answer

(B) Let the roots be $\alpha, \beta, \gamma, \delta$ such that $\alpha+\beta=0$,which implies $\beta=-\alpha$. Since $\alpha$ and $-\alpha$ are roots,the polynomial is divisible by $(x-\alpha)(x+\alpha) = x^2-\alpha^2$. Let the other two roots be $\gamma$ and $\delta$. Then $(x^2-\alpha^2)(x^2-Sx+P) = x^4-Sx^3+(P-\alpha^2)x^2+S\alpha^2x-P\alpha^2 = x^4-x^3-16x^2+4x+48$. Comparing coefficients: $S=1$ $P-\alpha^2=-16$ $S\alpha^2=4$ $\Rightarrow 1 \cdot \alpha^2=4$ $\Rightarrow \alpha^2=4$. Thus,$\alpha=2, \beta=-2$. From $P-\alpha^2=-16$,we get $P-4=-16 \Rightarrow P=-12$. Also,$-P\alpha^2=48 \Rightarrow -(-12)(4)=48$,which is consistent. Since $S=\gamma+\delta=1$ and $P=\gamma\delta=-12$,we have $\gamma^2+\delta^2 = (\gamma+\delta)^2-2\gamma\delta = 1^2-2(-12) = 1+24=25$. Then $\gamma^4+\delta^4 = (\gamma^2+\delta^2)^2-2\gamma^2\delta^2 = 25^2-2(-12)^2 = 625-288=337$. Finally,$\alpha^4+\beta^4+\gamma^4+\delta^4 = 2^4+(-2)^4+337 = 16+16+337 = 369$.

The sum of two roots of the equation $x^4-x^3-16x^2+4x+48=0$ is zero. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation,then $\alpha^4+\beta^4+\gamma^4+\delta^4=$

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