Solving equations with multiple roots So I was bored and tackled a bonus exercise during math class, and I managed to derive that the solution could be found as the solution for $x$ in the following equation:
$$\frac{\sqrt{x^2-9}+3\sqrt{x^2-1}}{x^2}-\frac12\sqrt{3}=0$$
Using my Graphing calculator, I find that $x\approx4.16331999$. The square of that is approximately $17.33333333$, so I suspect that the value of $x$ should be $\sqrt{17\frac13}$. I could substitute the $x$ in the equation above to verify this, but I'm not satisfied with that.
Instead, I was wondering if there was a way to solve the equation for $x$ instead of just making a guess using a calculator. However, I don't know how to solve the equation since because I can't seem to get rid of the square roots.
I am aware that there might be another way to solve the original problem, but it bothers me that I don't know how to solve the equation above even though an algebraic solution seems to exist.
How would I solve an equation like the one above without using a calculator?
P.S.
For those interested, the original problem was the following:
An equilateral triangular flag is hung between two poles, one with height 4 and one with height 3. Two corners of the flag are attached to tops of the poles, and the remaining corner touches the ground. What is the length of the sides of the flag?
It was the last and hardest question in the "Wiskunde Olympiade" (Math Olympics) of 2007, and 0% of the participants answered correctly.
 A: The equation you want to develop being $$\frac{\sqrt{x^2-9}+3\sqrt{x^2-1}}{x^2}-\frac12\sqrt{3}=0$$ start setting $x^2=t$ as suggested by  Shubham; so it becomes $$\frac{\sqrt{t-9}+3\sqrt{t-1}}{t}-\frac12\sqrt{3}=0$$ Multiply everything by $t$ and move the second part to the rhs. The expression becomes $$\sqrt{t-9}+3\sqrt{t-1}=\frac {\sqrt{3}}{2}t$$ Let us define $a=\sqrt{t-9}$,$b=3\sqrt{t-1}$,$c=\frac {\sqrt{3}}{2}t$. So we have now $$a+b=c$$ Square both sides and develop, moving $a^2$ and $b^2$ to the rhs. So $$2ab=c^2-a^2-b^2$$ and square again for reaching finally $$4a^2b^2=(c^2-a^2-b^2)^2$$ and now replace. Before simplification, the result is $$-\frac{9 t^4}{16}+15 t^3-91 t^2=0$$ $t^2$ can be factored and you end with $$(-\frac{9 t^2}{16}+15 t-91) t^2=0$$ The roots of the quadratic equation are $t=\frac{28}{3}$ and $t=\frac{52}{3}$. We must discard $t=0$ and $t=\frac{28}{3}$ because of the domain definition. So, the only root left is $t=\frac{52}{3}$ and, since $t=x^2$, the solutions are $$x_{\pm}=\pm \sqrt \frac{52}{3}$$ Just be patient.
