If alpha+beta=-2 and alpha^3+beta^3=-56,then | vedclass

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(D) Given that,$\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$.We know that $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta) = -56$.Subst

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$x^2+2x-8=0$

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(D) Given that,$\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$.We know that $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta) = -56$.Substituting $\alpha+\beta = -2$,we get $-2(\alpha^2+\beta^2-\alpha\beta) = -56$,which implies $\alpha^2+\beta^2-\alpha\beta = 28$.Also,$(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta = (-2)^2 = 4$.Subtracting the two equations: $(\alpha^2+\beta^2+2\alpha\beta) - (\alpha^2+\beta^2-\alpha\beta) = 4 - 28$.$3\alpha\beta = -24$,so $\alpha\beta = -8$.The quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha+\beta)x + \alpha\beta = 0$.Substituting the values,$x^2 - (-2)x + (-8) = 0$,which simplifies to $x^2+2x-8=0$.

If $\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$,then the quadratic equation whose roots are $\alpha$ and $\beta$ is

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