# Where did the $\pm$ go?

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$$Second Method:- Let the roots of equation (1) be \alpha and \beta, then \alpha+\beta=-\dfrac{c}{a} and \alpha\beta=\dfrac{c}{a} It is given that \dfrac{\alpha}{\beta}=\dfrac{p}{q}. Now, \text{LHS}=\sqrt{\dfrac{\alpha}{\beta}}+\sqrt{\dfrac{\beta}{\alpha}}+\sqrt{\dfrac{c}{a}}=\dfrac{\alpha+\beta}{\sqrt{\alpha\beta}}=\dfrac{-c/a}{\sqrt{c/a}}+\sqrt{\dfrac{c}{a}}=-\sqrt{\dfrac{c}{a}}+\sqrt{\dfrac{c}{a}}=0=\text{RHS} My attempt at a solution:- Its pretty much same as the first method given in the book but the only difference comes at the point where the book writes \alpha=\sqrt{\dfrac{c}{a}}\cdot\sqrt{\dfrac{1}{pq}}, instead of this shouldn't it be \alpha=\pm\sqrt{\dfrac{c}{a}}\cdot\sqrt{\dfrac{1}{pq}}, which finally resulted in \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}\pm\sqrt{\dfrac{c}{a}}=0. But testing the final result that I got like that done in the second method in the book only the expression that is to be proved was 0 but not the other one, i.e. \sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}-\sqrt{\dfrac{c}{a}}=0 this one. So, what is the reason that the book outright did not consider the other case without any reasoning? ## 1 Answer If the roots are r and s then$${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\$.

• 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. – user350331 Nov 6 '16 at 12:15