How to solve an equation with lots of square roots? Is there any easier way to solve the following equation? I've tried to solve with pen and paper. But after consuming lot of time and paper I am so tired that I am posting it here.
$$h=\dfrac{\sqrt{w^2+x^2}\sqrt{y^2-w^2}}{\sqrt{w^2+x^2}+\sqrt{y^2-w^2}}$$
And I need to solve the equation with respect to $w$.
Any help is appreciated. Thanks in advance.
 A: let $a = \sqrt{w^2+x^2}$ and $b = \sqrt {y^2-w^2}$.
You know that $h = ab/(a+b)$ and $a^2+b^2 = w^2+y^2$.
$(a+b)^2 = a^2+b^2+2ab = (w^2+y^2)+2h(a+b)$, so $((a+b)-h)^2 = w^2+y^2+h^2$, and $(a+b) = h \pm \sqrt{w^2+y^2+h^2}$.
With this you have two possibilities for the pair $(a+b,ab)$, which gives you two possible pairs for $\{a,b\}$ by solving the corresponding degree two equations.
Finally, once you know $a$ or $b$, you have $w = \pm \sqrt {a^2-x^2} = \pm \sqrt {y^2-b^2}$. This should give you at most $8$ possible values for $w$.
A: Start by multiplying by the conjugate:  $$h=\dfrac{\sqrt{w^2+x^2}\sqrt{y^2-w^2}}{\sqrt{w^2+x^2}+\sqrt{y^2-w^2}}=\dfrac{\sqrt{w^2+x^2}\sqrt{y^2-w^2}}{\sqrt{w^2+x^2}+\sqrt{y^2-w^2}}\cdot\dfrac{\sqrt{w^2+x^2}-\sqrt{y^2-w^2}}{\sqrt{w^2+x^2}-\sqrt{y^2-w^2}}\\ \frac {(w^2+x^2)\sqrt{y^2-w^2}-(y^2-w^2)\sqrt{w^2+x^2}}{x^2+y^2}$$ and we are down two.  Now the usual technique of isolate one term, square, isolate the remaining square root, square again will get rid of the remaining ones.  You may have a mess, or things may cancel nicely.  Note that your equation is in $w^2$, so the degree is only half of what it looks like.
