Let p and q be real numbers such that p neq 0 | vedclass

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(B) Given $\alpha + \beta = -p$ and $\alpha^3 + \beta^3 = q$.Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta

Vedclass · Accepted Answer

$(p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0$

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(B) Given $\alpha + \beta = -p$ and $\alpha^3 + \beta^3 = q$.Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$,we have:$q = (-p)^3 - 3\alpha\beta(-p) = -p^3 + 3p\alpha\beta$.Thus,$3p\alpha\beta = p^3 + q$,which implies $\alpha\beta = \frac{p^3 + q}{3p}$.The roots of the required quadratic equation are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.Sum of roots $= \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$.Substituting the values: $\frac{(-p)^2 - 2(\frac{p^3 + q}{3p})}{\frac{p^3 + q}{3p}} = \frac{p^2 - \frac{2(p^3 + q)}{3p}}{\frac{p^3 + q}{3p}} = \frac{3p^3 - 2p^3 - 2q}{p^3 + q} = \frac{p^3 - 2q}{p^3 + q}$.Product of roots $= \frac{\alpha}{\beta} 	imes \frac{\beta}{\alpha} = 1$.The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \left(\frac{p^3 - 2q}{p^3 + q}ight)x + 1 = 0$.Multiplying by $(p^3 + q)$,we get $(p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0$.

Let $p$ and $q$ be real numbers such that $p \neq 0, p^3 \neq q$ and $p^3 \neq -q$. If $\alpha$ and $\beta$ are non-zero numbers satisfying $\alpha + \beta = -p$ and $\alpha^3 + \beta^3 = q$,then find the quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.

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