If $\alpha $ and $\beta $ are the roots of $px^2 + qx + q=0$,

Question

If $\alpha $ and $\beta $ are the roots of $px^2 + qx + q=0$, prove that: $$\sqrt {\dfrac {\alpha}{\beta}} + \sqrt {\dfrac {\beta }{\alpha }} + \sqrt {\dfrac {q}{p}}=0$$.

My Attempt:

$$px^2+qx+q=0$$ Then, $$\alpha + \beta = \dfrac {-q}{p}$$ $$\alpha. \beta=\dfrac {q}{p}$$

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Answer 1

hint: You prove: $\left(\sqrt{\dfrac{\alpha}{\beta}} + \sqrt{\dfrac{\beta}{\alpha}}\right)^2 = \left(-\sqrt{\dfrac{p}{q}}\right)^2 \iff \dfrac{(\alpha+\beta)^2- 2\alpha\beta}{\alpha\beta} + 2 = \dfrac{p}{q}$. Can you take it from here?

Answer 2

$\left(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}\right)\left(\sqrt{\frac{\alpha}{\beta}}-\sqrt{\frac{\beta}{\alpha}}\right)=\frac{\alpha}{\beta}-\frac{\beta}{\alpha}=\frac{\alpha^2-\beta^2}{\alpha\beta}=\frac{(\alpha+\beta)(\alpha-\beta)}{\alpha\beta}=\frac{-q/p(\alpha-\beta)}{q/p}$

Now, $\left(\sqrt{\frac{\alpha}{\beta}}-\sqrt{\frac{\beta}{\alpha}}\right)=\frac{(\alpha/\beta)-1}{\sqrt{\alpha/\beta}}=\frac{\alpha-\beta}{\beta\cdot \sqrt{\alpha/\beta}}$

$\therefore \left(\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}\right)=-\beta\cdot \sqrt{\frac{\alpha}{\beta}}=-\sqrt{\alpha\beta}=-\sqrt{\frac{q}{p}}$

$\therefore \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{q}{p}}=0$

Answer 3

(Followup on previous comment, but too long to post as another comment.)

The conclusion does not hold true in general. Consider for example the case $\,p=q=1\,$, where the roots of $\,x^2+x+1=0\,$ are in fact the cube roots of unity:

$$ x^2+x+1=0 \;\;\implies\;\; \alpha=\frac{-1+i \sqrt{3}}{2}\,,\;\beta=\frac{-1-i\sqrt{3}}{2} = \frac{1}{\alpha}=\alpha^2\ $$

Then $\,\displaystyle \frac{\alpha}{\beta}=\frac{1}{\alpha}=\beta\,$ and $\displaystyle\,\frac{\beta}{\alpha} = \alpha\,$. Assuming that $\,\sqrt{\,\cdot\,}\,$ denotes the principal value of the complex square root (which is one of the "untold assumptions" that my first comment asked about, still unanswered), $\displaystyle \,\sqrt{\frac{-1 \pm i \sqrt{3}}{2}} = \frac{1 \pm i \sqrt{3}}{2}\,$, and it follows that:

$$ \sqrt {\dfrac {\alpha}{\beta}} + \sqrt {\dfrac {\beta }{\alpha }} + \sqrt {\dfrac {q}{p}}=\sqrt{\beta}+\sqrt{\alpha}+1=\frac{1 - i \sqrt{3}}{2}+\frac{1 + i \sqrt{3}}{2}+1 = 2 \ne 0 $$