# Equating sums of square roots

I solved the following equation the hard way: $$\sqrt{x+1} +\sqrt{x+33}=\sqrt{x+6} +\sqrt{x+22}$$ The only solution is $x=3$. I am wondering if there is some easy observation that solves the equation without squaring both sides?

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Squaring is your friend in these types of problems. –  diimension Nov 13 '12 at 9:06
Do you mean the only integer solution is $x=3$? –  TonyK Nov 13 '12 at 10:45
Depending on the source, you might guess that there is an integral solution and trying some small ones will find $3$. Then you can take the derivative of the difference of the two sides and find that it is of constant sign to show there are no more real solutions. –  Ross Millikan Nov 19 '12 at 20:10
Please transfer your acceptance from my wrong, now deleted, answer to Matthew Conroy's correct answer. –  John Bentin Nov 22 '12 at 10:54

In general, if an equation of the form $$\sqrt{x+a}+\sqrt{x+b} = \sqrt{x+c} + \sqrt{x+d}$$ has a solution, that solution is $$x = \frac{t^2-16s^2ab}{8s(2s(a+b)+t)}$$ where $s=a+b-c-d$ and $t=-s^2-4ab+4cd$. A bit messy, but you can be sure that the equation has, at most, one solution. If you find a small integer solution that works, then you are done.

To prove this general solution, you just square repeatedly and judiciously, simplifying as you go so you avoid getting any equation more complex than a quadratic at any stage (and those simplify to linear, and so we get only a single solution).

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You are right. My revised answer may still be wrong, so I would be grateful if you would check it. –  John Bentin Nov 22 '12 at 11:36
@JohnBentin I tried it out with some complicated examples, and your solution appears to be correct. Cheers! –  Matthew Conroy Nov 25 '12 at 1:17

Edit: My answer was wrong. Please refer to Matthew Conroy's correct answer instead. My version of his solution is $$x=\dfrac{(4s_2-s_1^2)^2-64s_4}{64s_3-8s_1(4s_2-s_1^2)},$$where $s_1, s_2, s_3, s_4$ are respectively the cubic, quadratic, linear, and constant coefficients of the polynomial expansion of $(x+a)(x+b)(x+c)(x+d)$.

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It is not quite as complicated as you describe. See my answer. Cheers! –  Matthew Conroy Nov 20 '12 at 23:29