A polynomial problem related to lx^2 + nx + n If the roots of  $lx^2 + nx + n = 0$ are in the ratio $p:q$, find the value of $\sqrt{\frac{p}{q}}$ + $\sqrt{\frac{q}{p}}$ + $\sqrt{\frac{n}{l}}$.
How to go about this problem?

Let $x_1$ and $x_2$ be the roots. we know that $x_1x_2=\frac n l$ or that $x_1=\frac{1}{x_2} \frac nl$. Since we have to calculate $\sqrt{\frac n l}$ then $n$ and $l$ should have the same sign. Let them both be positive. 
Now if $\frac{x_1}{x_2}=\frac pq$ then we have that $\frac p q=\frac 1 {x^2} \frac n l$ therefore
\begin{align}
\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}}&=\sqrt{\frac 1 {x_2^2} \frac n l} + \sqrt{x_2^2 \frac l n} + \sqrt{\frac{n}{l}}\\
&=\sqrt{\frac{n}{l}}\Big(\sqrt{\frac 1 {x_2^2} } + \frac l n\sqrt{x_2^2 } +1\Big)\\
\end{align}
now we should argue about the sign of $x_2$. If the roots are real and $n$ and $l$ are of the same sign, then both roots should either be negative of positive. Since we assumed that $n>0$, both roots will be negative. Hence
\begin{align}
\sqrt{\frac 1 {x_2^2} } + \frac l n\sqrt{x_2^2 }&=-\frac 1 {x_2}  -\frac 1 {x_1}\\
&=1
\end{align}
you could check for yourself what happens if both $n$ and $l$ are negative :-)
