If frac{1}{sqrt{alpha}} and frac{1}{sqrt{beta | vedclass

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(A) Given that $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of $ax^2 + bx + 1 = 0$.From the sum of roots: $\frac{1}{\sqrt{\alp

Vedclass · Accepted Answer

$\alpha^{3/2}$ and $\beta^{3/2}$

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(A) Given that $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of $ax^2 + bx + 1 = 0$.From the sum of roots: $\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a} \Rightarrow \frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha\beta}} = -\frac{b}{a}$.From the product of roots: $\frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a} \Rightarrow a = \sqrt{\alpha\beta}$.Substituting $a$ in the sum of roots equation: $\frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha\beta}} = -\frac{b}{\sqrt{\alpha\beta}} \Rightarrow b = -(\sqrt{\alpha} + \sqrt{\beta})$.The given equation is $x^2 + (b^3 - 3ab)x + a^3 = 0$.Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$,we have $b^3 = -(\sqrt{\alpha} + \sqrt{\beta})^3 = -(\alpha^{3/2} + \beta^{3/2} + 3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta}))$.Also,$3ab = 3(\sqrt{\alpha\beta})(-(\sqrt{\alpha} + \sqrt{\beta})) = -3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta})$.Thus,$b^3 - 3ab = -(\alpha^{3/2} + \beta^{3/2} + 3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta})) + 3\sqrt{\alpha\beta}(\sqrt{\alpha} + \sqrt{\beta}) = -(\alpha^{3/2} + \beta^{3/2})$.And $a^3 = (\sqrt{\alpha\beta})^3 = \alpha^{3/2}\beta^{3/2}$.The equation becomes $x^2 - (\alpha^{3/2} + \beta^{3/2})x + \alpha^{3/2}\beta^{3/2} = 0$.The roots of this quadratic equation are $\alpha^{3/2}$ and $\beta^{3/2}$.

If $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of the equation $ax^2 + bx + 1 = 0$ $(a \neq 0, a, b \in R)$,then the equation $x(x + b^3) + (a^3 - 3abx) = 0$ has roots:

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