If (alpha+sqrt{beta}) and (alpha-sqrt{beta}) | vedclass

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(A) Given that $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are roots of $x^{2}+px+q=0$.Sum of roots: $(\alpha+\sqrt{\beta}) + (\alpha-\sqrt{\b

Vedclass · Accepted Answer

$(\frac{1}{\alpha}+\frac{1}{\sqrt{\beta}})$ and $(\frac{1}{\alpha}-\frac{1}{\sqrt{\beta}})$

Vedclass · Answer

(A) Given that $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are roots of $x^{2}+px+q=0$.Sum of roots: $(\alpha+\sqrt{\beta}) + (\alpha-\sqrt{\beta}) = -p \Rightarrow 2\alpha = -p \Rightarrow \alpha = -\frac{p}{2}$.Product of roots: $(\alpha+\sqrt{\beta})(\alpha-\sqrt{\beta}) = q \Rightarrow \alpha^{2}-\beta = q \Rightarrow \beta = \alpha^{2}-q = \frac{p^{2}}{4}-q$.Thus, $p^{2}-4q = 4\beta$.Substitute into the equation $(p^{2}-4q)(p^{2}x^{2}+4px)-16q=0$:$4\beta(p^{2}x^{2}+4px) - 16q = 0$.Since $p = -2\alpha$, $p^{2} = 4\alpha^{2}$ and $q = \alpha^{2}-\beta$:$4\beta(4\alpha^{2}x^{2}-8\alpha x) - 16(\alpha^{2}-\beta) = 0$.Divide by $4$:$\beta(\alpha^{2}x^{2}-2\alpha x) - (\alpha^{2}-\beta) = 0$.$\alpha^{2}\beta x^{2} - 2\alpha\beta x - \alpha^{2} + \beta = 0$.Rearranging terms: $\alpha^{2}(\beta x^{2}-1) - \beta(2\alpha x - 1) = 0$ is not straightforward, so let's simplify $4\beta(p^{2}x^{2}+4px) = 16q$:$4\beta(4\alpha^{2}x^{2}-8\alpha x) = 16(\alpha^{2}-\beta) \Rightarrow \beta(\alpha^{2}x^{2}-2\alpha x) = \alpha^{2}-\beta$.$\alpha^{2}\beta x^{2} - 2\alpha\beta x + \beta = \alpha^{2} \Rightarrow \beta(\alpha x - 1)^{2} = \alpha^{2}$.$(\alpha x - 1)^{2} = \frac{\alpha^{2}}{\beta} \Rightarrow \alpha x - 1 = \pm \frac{\alpha}{\sqrt{\beta}}$.$\alpha x = 1 \pm \frac{\alpha}{\sqrt{\beta}} \Rightarrow x = \frac{1}{\alpha} \pm \frac{1}{\sqrt{\beta}}$.

If $(\alpha+\sqrt{\beta})$ and $(\alpha-\sqrt{\beta})$ are the roots of the equation $x^{2}+px+q=0$, where $\alpha, \beta, p$ and $q$ are real, then the roots of the equation $(p^{2}-4q)(p^{2}x^{2}+4px)-16q=0$ are

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