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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.

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  • $\begingroup$ Whoops. Thanks Ross; I've fixed it. $\endgroup$ – Sujaan Kunalan Oct 3 '13 at 16:04
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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.

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    $\begingroup$ I tried some ideas: they all look as nasty as a nightmare after eating too many greasy hamburgers. Either there's some hidden trick somewhere and I missed it or this is an exercise from hell. $\endgroup$ – DonAntonio Oct 3 '13 at 16:12
  • $\begingroup$ Thanks a lot for your help. I don't what I'm gonna do, all those papers in my notebook are still scaring me!! $\endgroup$ – maksbd19 Oct 3 '13 at 16:15
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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 $w$.

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  • $\begingroup$ That's a great answer, but I'm sorry that I have already accepted one. So that I only have to vote your answer up. Thanks for your help. $\endgroup$ – maksbd19 Oct 4 '13 at 14:12

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