well first you can move one of the square roots to the right hand side and then square both sides, then you can issolate the remaining square root to one side of the equation and square again. Finally after moving all variables to one side of the equation you end up with the nice 8 degree polynomial mentioned by Oskar. That polynomial looks like this in its factored form
so obviously we have x=2 as a solution with multiplicity 2
and it can easily be shown with most graphing calculators that all the remaining zeros of this polynomial are complex, thus the only real solution is x=2.
Now for those that are curious the complex zeros are
Posted by Daniel
on 2006-08-16 16:21:36