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If $\alpha, \beta$ are the roots of $a x^{2}+b x+b=0$, then what is
$\frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\frac{\sqrt{b}}{\sqrt{a}}$ equal to?
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$\frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\frac{\sqrt{b}}{\sqrt{a}}$ equal to?
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Let $\alpha$ and $\beta$ are the roots of $a x^{2}+b x+b=0$
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}$ and $\alpha \beta=\frac{\mathrm{b}}{\mathrm{a}}$
Consider, $\frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\frac{\sqrt{\mathrm{b}}}{\sqrt{a}}=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\frac{\sqrt{\mathrm{b}}}{\sqrt{a}}$
$=\frac{-b / a}{\sqrt{b / a}}+\sqrt{\frac{b}{a}}=-\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{a}}$ (by rationalizing)
$\alpha+\beta=-\frac{\mathrm{b}}{\mathrm{a}}$ and $\alpha \beta=\frac{\mathrm{b}}{\mathrm{a}}$
Consider, $\frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}}+\frac{\sqrt{\mathrm{b}}}{\sqrt{a}}=\frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\frac{\sqrt{\mathrm{b}}}{\sqrt{a}}$
$=\frac{-b / a}{\sqrt{b / a}}+\sqrt{\frac{b}{a}}=-\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{a}}$ (by rationalizing)
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