If the sum of two roots alpha, beta of the eq | vedclass

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(D) Let $\alpha, \beta, \gamma, \delta$ be the roots of $x^4-x^3-8 x^2+2 x+12=0$. Given $\alpha+\beta=0$. We can write the polynomial as $(x^2+a)(x^2-

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$5$

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(D) Let $\alpha, \beta, \gamma, \delta$ be the roots of $x^4-x^3-8 x^2+2 x+12=0$. Given $\alpha+\beta=0$. We can write the polynomial as $(x^2+a)(x^2-x+b) = x^4-x^3+(a+b)x^2-ax+ab$. Comparing the coefficients with $x^4-x^3-8x^2+2x+12=0$: $-a=2 \implies a=-2$. $ab=12 \implies -2b=12 \implies b=-6$. Thus,$x^4-x^3-8x^2+2x+12 = (x^2-2)(x^2-x-6) = (x^2-2)(x-3)(x+2)$. The roots are $\pm\sqrt{2}, 3, -2$. Since $\alpha+\beta=0$,we have $\alpha=\sqrt{2}, \beta=-\sqrt{2}$. The other roots are $\gamma=3, \delta=-2$ (given $\gamma > \delta$). Therefore,$3\gamma+2\delta = 3(3)+2(-2) = 9-4 = 5$.

If the sum of two roots $\alpha, \beta$ of the equation $x^4-x^3-8 x^2+2 x+12=0$ is zero and $\gamma, \delta$ $(\gamma > \delta)$ are its other roots,then $3 \gamma+2 \delta=$

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