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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?

Later
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Adam
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    Squaring is your friend in these types of problems. – diimension Nov 13 '12 at 09:06
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    Please use more informative titles. – Did Nov 13 '12 at 10:18
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    Do you mean the only integer solution is $x=3$? – TonyK Nov 13 '12 at 10:45
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    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
  • Do not square! Two non-identical monotonius (either increasing or decreasing) functions can cross at no more than 1 point. It is easy to prove: 2nd crossing point will violate condition of functions being strictly increasing / decreasing. Here is my answer to a similar question. – Stoleg May 23 '13 at 22:35
  • @Stoleg I've converted your answer to a comment, since it is generally considered bad practice to give an answer which is just a link to another answer on the site. It's also worth noting that your other answer is incorrect. Consider for example $f(x)=x+2, g(x)=e^x$ which are both strictly increasing but cross twice. In general arbitrarily many crossings are possible. – Alex Becker May 23 '13 at 22:51

2 Answers2

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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).

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

John Bentin
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