# 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?

• 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.
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 $(x+a)(x+b)(x+c)(x+d)$.