An algebra problem by milind prabhu

$-\sqrt{x}=x\\ x+\sqrt{x}=0\\ \sqrt{x}\left(\sqrt{x}+1\right)=0$

$\sqrt{x}=0 \implies x=0$

$\sqrt{x}+1=0 \implies \sqrt{x}=-1 \implies$ No real solution

Therefore, the only real solution of this equation is $x=0$ . There are $\boxed{0}$ non-zero real solutions for this equation

For the left hand to be defined, $x$ has to be positive. The $\sqrt { x }$ is always positive or zero. Here it cannot be zero so it is positive. So the left hand side is always negative. The right hand side on the other hand is always positive as $x$ is always positive. So there exists no non zero real solution to the equation.

For any $x$ , its squared counterpart is always positive. Therefore, the answer is that there are 0 values that makes the equation true.

However, a more correct solution would be infinity because the question specified 'non-zero real number'. Since the square root of, let's say, 4 can be both 2 and - 2, the equation would be true for the negative root.

$\color{#D61F06}{\boxed{\sqrt{x}\geq 0}}\implies \color{#3D99F6}{\boxed{-\sqrt{x}=x\leq 0}}\\ \text{On the other hand:}\\ \color{#D61F06}{\sqrt{x}\geq 0}\implies \color{#20A900}{\boxed{x\geq 0}}\\ \color{#3D99F6}{x\leq 0},\color{#20A900}{x\geq 0}\implies \color{teal}{\boxed{\boxed{x=0}}}\\ \text{Therefore} \;x=0\;\text{is the only real solution to this equation}$

Another way is to sketch the graph. Clearly there are only one point of intersection (which is at the origin), and hence the number of non-zero solutions is 0.