Is it wrong to say that $100$ is solution of $\sqrt x+10=0$? 
Is it wrong to say that 100 is solution of $\sqrt x +10=0$?

I know that range of $\sqrt{x}$ is $[0, \infty)$ by convention. This convention is because of definition of a function. So if I consider $\sqrt x+10=0$ as a simple equation then can I say hundred is a solution to the equation?
I think I can say that because if I square both sides I get $x=100$ and also if I consider $\sqrt x+10=0$ just as a equation then there is no need of thinking about functions and the convention of omitting negative roots.  
 A: The solutions to $f(x) = k$ are a SUBSET of the solutions to $f(x)^2 = k^2$ but not all the solution to $f(x)^2 = k^2$ are solutions to $f(x) = k$.  Squaring both sides of an equation add extraneous solutions.
$\sqrt{x} + 10 = 0$
$\sqrt{x} = -10$ Doesn't just mean that $(\sqrt{x}^2 = (-10)^2$.  It ALSO means that $\sqrt{x} = -10 < 0$.
When we square both sides we LOSE information.
$\sqrt{x}^2 = (-10)^2$
$x = 100$ but we have completely LOST that $\sqrt{x} < 0$.
Any  $\sqrt{100} + 10 = 10 + 10 = 20 \ne 0$.  SO it simply DOESN'T work.
Consider this:
$x =2 $ has one solution. Square both sides and you get $x^2 = 4$.  Which has TWO solutions!!!  Where did that solution $x = -2$ come from?
It came because when we squared both sides we added invalid extraneous solutions.
Solutions to $x= 2$ is $\{2\}$.  Solutions to $x^2 = 4$ has solutions $\{2,-2\}$ and $\{2\} \subset \{2,-2\}$.  But it doesn't go the other way.  It only goes one way.
A: Consider the equation x^2 = 100
When we take the square root of the equation to solve it, we would write
x = ± sqrt(100)
Notice that there is a ± sign in front of the square root. The reason that is there is precisely because sqrt(x) is defined as the principle square root. 
So, when you evaluate sqrt(100) + 10, you get 10 + 10 = 20, proving your equation false.
A: $x=100$ is most definitely a solution to $\sqrt{x}-10=0$, because if you plug $x=100$ into the equation, you get $\sqrt{100}-10=0$, or $10-10=0$, which is true.
A: $√(x)+10=0$
$√(x)=-10$
$√(x)^2=(-10)^2$
$x=100$
$√(100)+10≠0$
Thus, since $√(100)+10$ does not equal zero, this equation has no solution. 
