# Why does $\sqrt x = -a$ have no solution?

Algebraically, can't you solve this equation by squaring both sides?: $$\sqrt x = -a$$ $$(\sqrt x)^2 = (-a)^2$$ $$x = a ^2$$

I can understand why $$x^2 = -a$$ has no solution, because a number multiplied by itself twice can't be a negative number. However, in the case above, $$-4$$ is technically still a square root of $$16$$.

• "-4 is techinally still a square root of 16" is incorrect. $\sqrt{b^2}=|b|$ Nov 26, 2020 at 6:18
• @RudyGoburt You need to provide us with a definition of $\sqrt x$.
– user838035
Nov 26, 2020 at 6:20
• Don't confuse the two square roots of a number, $\pm\sqrt a$, and the square root function $\sqrt a$.
– user65203
Nov 26, 2020 at 7:39

Usually when we write $$\sqrt{x}$$ where $$x$$ is a real number, we usually mean the principal root. This means the positive square root. While $$(-4)^2=16$$, we don't say that $$\sqrt{16}=-4$$, because $$\sqrt{16}=4$$, and if $$\sqrt{16}$$ were both values, it wouldn't be a function. This is why when squaring an equation can sometimes lead to extraneous solutions.

When taking the square root of an equation, we attach a $$\pm$$. So if $$a^2=16$$, we do $$a=\pm\sqrt{16}=4,-4$$.

In the reals, if $$a>0$$ then $$\sqrt x=-a$$ has no solution since the square root in this context is usually understood to return a non-negative number.

In the complexes, $$\sqrt x=-a$$ has two solutions as expected: $$x=\pm i\sqrt a$$.

• So it's because the positive solution is used more frequently? Nov 26, 2020 at 6:24
• @RudyGoburt Non-negative solution is understood by convention. Nov 26, 2020 at 6:24

One more way to see why your argument is not valid,

Assuming $$a$$ is positive, Define the function $$f: [0,\infty) \to [0, \infty)$$ by $$f(x) = \sqrt x$$ and generally when we use $$\sqrt x$$ we mean this function $$f$$.

So $$\sqrt x$$ is a unique,positive value. By definition of $$\sqrt x$$, it cannot take negative values, hence the equation provided by you does not have a solution.