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(D) Given the quadratic equation $x^{2}+p x+q=0$,the sum of roots $\alpha+\beta = -p$ and the product of roots $\alpha \beta = q$. First,calculate $\a

Vedclass · Accepted Answer

$3 p q-p^{3}$ and $(p^{2}-q)(p^{2}-3 q)$

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(D) Given the quadratic equation $x^{2}+p x+q=0$,the sum of roots $\alpha+\beta = -p$ and the product of roots $\alpha \beta = q$. First,calculate $\alpha^{3}+\beta^{3}$: $\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta) = (-p)^{3}-3 q(-p) = -p^{3}+3 p q = 3 p q-p^{3}$. Next,calculate $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}$: $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2} - \alpha^{2} \beta^{2} = [(\alpha+\beta)^{2}-2 \alpha \beta]^{2} - (\alpha \beta)^{2} = [(-p)^{2}-2 q]^{2} - q^{2} = (p^{2}-2 q)^{2} - q^{2} = p^{4}-4 p^{2} q+4 q^{2}-q^{2} = p^{4}-4 p^{2} q+3 q^{2} = p^{2}(p^{2}-3 q)-q(p^{2}-3 q) = (p^{2}-q)(p^{2}-3 q)$. Thus,the values are $3 p q-p^{3}$ and $(p^{2}-q)(p^{2}-3 q)$.

If $\alpha, \beta$ are the roots of the quadratic equation $x^{2}+p x+q=0,$ then the values of $\alpha^{3}+\beta^{3}$ and $\alpha^{4}+\alpha^{2} \beta^{2}+\beta^{4}$ are respectively

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