If the sum of two roots of the equation x^4+2 | vedclass

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(B) Let the roots of the equation $x^4+2x^3-7x^2-8x+12=0$ be $\alpha, \beta, \gamma, \delta$. Given that the sum of two roots is zero,let $\alpha + \b

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$10$

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(B) Let the roots of the equation $x^4+2x^3-7x^2-8x+12=0$ be $\alpha, \beta, \gamma, \delta$. Given that the sum of two roots is zero,let $\alpha + \beta = 0$,which implies $\beta = -\alpha$. From Vieta's formulas,the sum of roots is $\alpha + \beta + \gamma + \delta = -2$. Since $\alpha + \beta = 0$,we have $\gamma + \delta = -2$. The product of roots is $\alpha \beta \gamma \delta = 12$. Since $\beta = -\alpha$,we have $-\alpha^2 \gamma \delta = 12$,or $\alpha^2 \gamma \delta = -12$. Also,the sum of roots taken two at a time is $\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = -7$. Substituting $\beta = -\alpha$,we get $-\alpha^2 + \alpha(\gamma + \delta) - \alpha(\gamma + \delta) + \gamma \delta = -7$. This simplifies to $-\alpha^2 + \gamma \delta = -7$,so $\gamma \delta = \alpha^2 - 7$. Substituting $\gamma \delta$ into the product equation: $\alpha^2(\alpha^2 - 7) = -12$,which gives $\alpha^4 - 7\alpha^2 + 12 = 0$. Let $y = \alpha^2$,then $y^2 - 7y + 12 = 0$,so $(y-3)(y-4) = 0$. Thus $\alpha^2 = 3$ or $\alpha^2 = 4$. If $\alpha^2 = 4$,then $\gamma \delta = 4 - 7 = -3$. We know $\gamma + \delta = -2$. The sum of squares $\gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma \delta = (-2)^2 - 2(-3) = 4 + 6 = 10$. If $\alpha^2 = 3$,then $\gamma \delta = 3 - 7 = -4$. Then $\gamma^2 + \delta^2 = (-2)^2 - 2(-4) = 4 + 8 = 12$. Checking the original equation,the roots are $2, -2, 1, -3$. The sum of two roots being zero is $2 + (-2) = 0$. The other two roots are $1$ and $-3$. The sum of their squares is $1^2 + (-3)^2 = 1 + 9 = 10$.

If the sum of two roots of the equation $x^4+2x^3-7x^2-8x+12=0$ is zero,then the sum of the squares of the other two roots is

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