Question

If the ratio of the roots of the equation $l{x}^2+nx+n=0$ is equal to $p:q$, prove that $\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0$

Answer 1

Let $\alpha$ and $\beta$ be two roots of $l{x}^2+nx+n=0$

$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=\frac{\sqrt{p}}{\sqrt{q}}+\frac{\sqrt{q}}{\sqrt{p}}+\frac{\sqrt{n}}{\sqrt{l}}$

$=\frac{p+q}{\sqrt{p}+\sqrt{q}}+\frac{\sqrt{n}}{\sqrt{l}}-----\left(1\right)$

Given, $\alpha :\beta =p:q$

Let $\alpha =px$ and $\beta =qx$ (where $x=$ constant of proportionality).

$\alpha +\beta =-\frac{n}{l}$

Or, $px+qx=-\frac{n}{l}$

Or, $x(p+q)=-\frac{n}{l}\Rightarrow p+q=-\frac{n}{lx}-----\left(2\right)$

Again, $\alpha \beta =\frac{n}{l}$

Or, $px.qx=\frac{n}{l}$

Or, $pq{x}^2=\frac{n}{l}$

Or, $\sqrt{pq}x=\frac{\sqrt{n}}{\sqrt{l}}$

$\therefore \sqrt{pq}=\frac{\sqrt{n}}{\sqrt{l}x}-----\left(3\right)$

Now, from $\left(1\right)$

$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=\frac{-n\sqrt{l}x}{lx\sqrt{n}}+\frac{\sqrt{n}}{\sqrt{l}}$

$=-\frac{\sqrt{n}}{\sqrt{l}}+\frac{\sqrt{n}}{\sqrt{l}}=0$

$------proved$.