Solving the equation $\sqrt{x+1} +\sqrt{x+33}=\sqrt{x+6} +\sqrt{x+22}$ for $x$

Question

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?

Answer 1

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).

Answer 2

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 $(x+a)(x+b)(x+c)(x+d)$.