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 $ax^2 + bx + 1 = 0$. Sum of roots: $\frac{1}{\sqrt{\alpha}} + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given that $\frac{1}{\sqrt{\alpha}}$ and $\frac{1}{\sqrt{\beta}}$ are the roots of $ax^2 + bx + 1 = 0$. Sum of roots: $\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a} \Rightarrow \frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha \beta}} = -\frac{b}{a}$. Product of roots: $\frac{1}{\sqrt{\alpha \beta}} = \frac{1}{a} \Rightarrow a = \sqrt{\alpha \beta}$. Substituting $a$ into the sum equation: $\frac{\sqrt{\alpha} + \sqrt{\beta}}{\sqrt{\alpha \beta}} = -\frac{b}{\sqrt{\alpha \beta}} \Rightarrow b = -(\sqrt{\alpha} + \sqrt{\beta})$. The given equation is $x^2 + (b^3 - 3ab)x + a^3 = 0$. Note that $b^3 - 3ab = -(\sqrt{\alpha} + \sqrt{\beta})^3 - 3(\sqrt{\alpha \beta})(-(\sqrt{\alpha} + \sqrt{\beta})) = -(\alpha^{3/2} + \beta^{3/2} + 3\sqrt{\alpha \beta}(\sqrt{\alpha} + \sqrt{\beta})) + 3\sqrt{\alpha \beta}(\sqrt{\alpha} + \sqrt{\beta}) = -(\alpha^{3/2} + \beta^{3/2})$. Also,$a^3 = (\sqrt{\alpha \beta})^3 = \alpha^{3/2} \beta^{3/2}$. The equation becomes $x^2 - (\alpha^{3/2} + \beta^{3/2})x + \alpha^{3/2} \beta^{3/2} = 0$. This is a quadratic equation with roots $\alpha^{3/2}$ and $\beta^{3/2}$.

$A. \alpha + \beta + \gamma$	$(1) -\frac{43}{4}$
$B. \alpha^2 + \beta^2 + \gamma^2$	$(2) -\frac{7}{8}$
$C. \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$	$(3) 86$
$D. \frac{\alpha}{\beta \gamma} + \frac{\beta}{\gamma \alpha} + \frac{\gamma}{\alpha \beta}$	$(4) 0$
	$(5) 10$

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

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