Solution of an algebraic equation having unknown within modulus The question is this:
find the number of real values of $x$ for which $|x-3|+(x-3)^2+\sqrt{x-3}+|x+3|=0.$
Here is my attempt to answer:
Since it's a real equation and $\sqrt{x-3}$ is present here, therefore we must have $x>3$. In this case this equation becomes 
$(x-3)+(x-3)^2+\sqrt{x-3}+(x+3)=0$
$\Rightarrow 2x+(x-3)^2+\sqrt{x-3}=0$
$\Rightarrow 2(x-3)+(x-3)^2+\sqrt{x-3}+6=0$
$\Rightarrow (x-3+1)^2+\sqrt{x-3}+5=0$
$\Rightarrow \{(x-2)^2+5\}^2=x-3$
$\Rightarrow (x-2)^4+10(x-2)^2-(x-2)+26=0,$ which is a biquadratic equation. 
I want to know whether I am going in the right direction. 

Hint (easier direction):
None of the summands is negative.
Could they all be zero?
