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

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(C) Let the roots of the equation $x^4-2x^3+x^2+4x-6=0$ be $\alpha, \beta, \gamma, \delta$. Given $\alpha + \beta = 0$. From the relation between root

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$-2$

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(C) Let the roots of the equation $x^4-2x^3+x^2+4x-6=0$ be $\alpha, \beta, \gamma, \delta$. Given $\alpha + \beta = 0$. From the relation between roots and coefficients: $\alpha + \beta + \gamma + \delta = -(-2)/1 = 2$. Since $\alpha + \beta = 0$,we have $\gamma + \delta = 2$. Also,the product of roots taken two at a time is $\alpha\beta + \gamma\delta + (\alpha+\beta)(\gamma+\delta) = 1$. Substituting $\alpha+\beta=0$ and $\gamma+\delta=2$,we get $\alpha\beta + \gamma\delta = 1$. From the product of roots taken three at a time: $\alpha\beta(\gamma+\delta) + \gamma\delta(\alpha+\beta) = -4$. Substituting $\alpha+\beta=0$ and $\gamma+\delta=2$,we get $2\alpha\beta = -4$,so $\alpha\beta = -2$. Then $\gamma\delta = 1 - (-2) = 3$. The sum of the squares of the other two roots is $\gamma^2 + \delta^2 = (\gamma+\delta)^2 - 2\gamma\delta$. Substituting the values,$\gamma^2 + \delta^2 = (2)^2 - 2(3) = 4 - 6 = -2$.

$A. \alpha + \beta + \gamma$	$(1) -\frac{43}{4}$
$B. \alpha^2 + \beta^2 + \gamma^2$	$(2) -\frac{7}{8}$
$C. \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$	$(3) 86$
$D. \frac{\alpha}{\beta \gamma} + \frac{\beta}{\gamma \alpha} + \frac{\gamma}{\alpha \beta}$	$(4) 0$
	$(5) 10$

If the sum of two roots of the equation $x^4-2x^3+x^2+4x-6=0$ is zero,then the sum of the squares of the other two roots is

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