# Solving $\frac{\sqrt {x}}{2}=-1$, I get $x=4$. Where did I go wrong? [duplicate]

We have the equation $$\frac{\sqrt {x}}{2}=-1$$ I proceed as follows $$\sqrt {x}=-2$$ $$x=4$$ Which does not certainly solve the equation. Where did I go wrong?

• Whenever you square both sides of an equation, you risk getting solutions to the new equation that don't satisfy the original. That's just how squaring works...It's an if-then kind of operation, not an only if kind.
– aman
Commented Mar 13, 2021 at 7:04
• This won't have any solution because there aren't any real numbers that can satisfy this equation. Commented Mar 13, 2021 at 7:08
• @Justin Where is my reasoning flawed?
– user893119
Commented Mar 13, 2021 at 7:16
• @PrasunBiswas So I cannot square both sides, while keeping the equation invariant?
– user893119
Commented Mar 13, 2021 at 7:28
• @PrasunBiswas They never teach us this in school. Anyways, thanks.
– user893119
Commented Mar 13, 2021 at 7:33

By definition of the square root :

• If $$\sqrt x=a$$, then

$$\begin{cases} x=a^2 \\ a≥0 \end {cases}$$

Now, you can check your solution.

Note that, $$-2$$ is the real-valued root of $$4$$. But, we must take only principal square root, which equals to $$2.$$

Small supplement:

• If $$\sqrt x=a$$, then you get $$x=a^2$$. This is correct. But, if the condition $$a≥0$$ doesn't hold, then the solution doesn't exist.
• Comments are not for extended discussion; this conversation has been moved to chat. Commented Mar 13, 2021 at 14:19