Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<
 A: 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 $\sqrt{8}$. 
A: 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 $10$.
A: Questions:


*

*Is the problem written correctly.

*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.
A: 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.
A: 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.
