# Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$

Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$ Little help here? >.<

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There are no real solutions, nor any complex solutions if you use the principal branch of the square root. Squaring both sides and simplifying gives you $20 \sqrt{x-4} = -92$.

EDIT: More generally, for any $a, b \ge 0$, $\sqrt{a + b} \le \sqrt{a} + \sqrt{b}$. Since $(x+4) - (x-4) = 8$, the most $\sqrt{x+4} - \sqrt{x-4}$ can be is $\sqrt{8}$.

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The inequality $\sqrt{a+b}< \sqrt{a}+\sqrt{b}$ hasa pretty obvious geometric proof, which leads to the following simple adaptation of your solution: draw the right angle triangle with edges $\sqrt{x-4}, \sqrt{x+4}, \sqrt{8}$, and then by the triangle inequality $\sqrt{x+4}< \sqrt{x-4}+\sqrt{8}$. – N. S. Sep 7 '12 at 2:39

We will assume that $x$ ranges over the reals $\ge 4$, to make sure that the square roots are real. Note that $$\sqrt{x+4}-\sqrt{x-4}=\frac{(\sqrt{x+4}-\sqrt{x-4})(\sqrt{x+4}+\sqrt{x-4})}{\sqrt{x+4}+\sqrt{x-4}} =\frac{8}{\sqrt{x+4}+\sqrt{x-4}} .$$ For $x\ge 4$, $\sqrt{x+4}+\sqrt{x-4}\ge 2\sqrt{2}$. It follows that $\sqrt{x+4}-\sqrt{x-4}\le \dfrac{8}{2\sqrt{2}}=2\sqrt{2}$ for all $x\ge 4$. In particular, $\sqrt{x+4}-\sqrt{x-4}$ cannot be equal to $10$.

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Questions:

1. Is the problem written correctly.

2. Are there restrictions on x?

Something does not seem right in the problem as posed.

Hint: Plot the left hand side and then plot the right hand side and see what it looks like.

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"No real solution" is surely an acceptable answer.... – Cameron Buie Sep 7 '12 at 0:16
That's possibly why the user asked this question. – abnry Sep 7 '12 at 1:23

Square both sides, and you get $$x - 4 + 20\sqrt{x - 4} + 100 = x + 4$$ This simplifies to $$20\sqrt{x - 4} = -92$$ or just $$\sqrt{x - 4} = -\frac{92}{20}$$ Since square roots of numbers are always nonnegative, this cannot have a solution.

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As others have said, there are no solutions within the usual rules. However, once we get to $\sqrt {x-4}=-4.6$ we can remember that square roots can be negative (despite the convention that $\sqrt x \ge 0$). Then we can square and add $4$to find $x=25.16$. Checking, we find $\sqrt {x+4}=5.4, \sqrt{x-4}=-4.6$ and the difference is truly $10$. You can decide if this is better than no answer at all.

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