Find solution set $\sqrt{x-5} + 5 =0$ Find solution set :
$\sqrt{x-5} + 5 =0$
Is there a solution to a set of complex numbers ?
 A: There are two square roots of any nonzero complex number.  Without further specification, it's not clear which is meant by $\sqrt{x-5}$.
If you allow both square roots, $x=30$ is the solution.  If you mean the principal branch of the square root, there are no solutions.
A: Yes, the solution here is $x = 30$.
The number $25$ which viewed as a complex number has two complex roots: $-5$ and $5$.
So if $x=30$, then $\sqrt{x-5} + 5 = 0$.  
Also note that $25$ is the only number which has a complex root of $-5$.
This proves $30$ is the only solution.   
This is some sort of synthetic solution.  
If you want an analytic solution then write $x=a+b.i$,
move the 5 to the right, raise to the power of 2 and see what happens.  
A: No.  There is no solution because $\sqrt{}$ is by definition non-negative.
However $\pm\sqrt{x-5} + 5 = 0$ does have a solution.
$\pm\sqrt{x-5} + 5 = 0$
$\pm\sqrt{x-5} = -5$
$(\pm\sqrt{x-5})^2 = (-5)^2$
$x-5 = 25$
$x = 30$.
But seriously.  You are NOT allowed to say $\sqrt{x-5} = -5$.  You just can't.  It's against the rules.
A: In order to  have a real solution the implied $\pm$ before the radical sign should have negative sign, $ x= 30 $ is the solution. Other solution is complex.
Note that for a complex solution one takes care of both signs automatically. Like in quadratic equation  $ x^2 - 8 x + 25 =0 \, $ the roots are $ 4 \pm 3 i $,
