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Question Statement:-

If the roots of the equation $ax^2+cx+c=0$ be in the ratio $p:q$, then show that $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{c}{a}}=0$


Solution provided by the book:-

Given equation is $ax^2+cx+c=0\tag{1}$

Let the roots be $p\alpha$ and $q\alpha$, then $(p+q)\alpha=-\dfrac{c}{a}\tag{2}$ $pq\alpha^2=\dfrac{c}{a}\tag{3}$

From $(3)$, $\alpha^2=\dfrac{c}{apq}$$\therefore \alpha=\sqrt{\dfrac{c}{a}}\cdot\dfrac{1}{\sqrt{pq}}\tag{4}$

Putting the value of $\alpha$ in $(2)$, we get

$$(p+q)\sqrt{\dfrac{c}{a}}\cdot\dfrac{1}{\sqrt{pq}}=\dfrac{-c}{a}\implies \dfrac{p+q}{\sqrt{pq}}=-\sqrt{\dfrac{c}{a}}\implies \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{c}{a}}=0$$

Second Method:- Let the roots of equation $(1)$ be $\alpha$ and $\beta$, then $\alpha+\beta=-\dfrac{c}{a}$ and $\alpha\beta=\dfrac{c}{a}$

It is given that $\dfrac{\alpha}{\beta}=\dfrac{p}{q}$.

Now, $\text{LHS}=\sqrt{\dfrac{\alpha}{\beta}}+\sqrt{\dfrac{\beta}{\alpha}}+\sqrt{\dfrac{c}{a}}=\dfrac{\alpha+\beta}{\sqrt{\alpha\beta}}=\dfrac{-c/a}{\sqrt{c/a}}+\sqrt{\dfrac{c}{a}}=-\sqrt{\dfrac{c}{a}}+\sqrt{\dfrac{c}{a}}=0=\text{RHS}$


My attempt at a solution:-

Its pretty much same as the first method given in the book but the only difference comes at the point where the book writes $\alpha=\sqrt{\dfrac{c}{a}}\cdot\sqrt{\dfrac{1}{pq}}$, instead of this shouldn't it be $\alpha=\pm\sqrt{\dfrac{c}{a}}\cdot\sqrt{\dfrac{1}{pq}}$, which finally resulted in $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}\pm\sqrt{\dfrac{c}{a}}=0$. But testing the final result that I got like that done in the second method in the book only the expression that is to be proved was $0$ but not the other one, i.e. $\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{c}{a}}=0$ this one.

So, what is the reason that the book outright did not consider the other case without any reasoning?

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If the roots are $r$ and $s$ then $${r\over s} = {p\over q} = {p\alpha \over q\alpha}.$$

If $\alpha$ were negative, we could just cancel the sign and make it positive. So we might as well assume it's positive. Any sign issues are absorbed into $p$ and $q$.

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  • $\begingroup$ That seems good enough. But your answer is a lot smaller than the work that I had to do to write the oh so long post...just kidding. $\endgroup$ – user350331 Nov 6 '16 at 12:15

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