Why can't √x = -1?

Question

So I had this problem: $x^2 + 9 + \sqrt{x^2 + 9} = 0$ which can be written like this $\sqrt{x^2 + 9}(\sqrt{x^2 + 9} + 1)$

And the obvious answer is $x = \pm3i$

but if you went deeper and tried $\sqrt{x^2 + 9} = -1$

you would get $x = \pm\sqrt{8}i$, which if we replace $x$ in the equation above gives the wrong answer.

So my question is why can't $\sqrt{x^2 + 9} = -1$ be true?

Answer 1

Aha. This is where my dilemma: $$\text{Is } x=1 \text { a solution of } x-2\sqrt x -3=0\text{ ?}$$

comes into full force. $\sqrt x$ is generally defined as the non-negative square root of a non-negative real number, thus, we would get $1-2-3=-4$, not a solution, but taking $\sqrt 1=-1$ we get $1+2-3=0$, a solution, but due to the definition of $\sqrt x$ the second is not accepted.

Let's solve this ourselves without plugging anything in. We see:

$$x-3=2\sqrt x\to 4x=(x^2-6x+9)\to(x^2-10x+9)=0$$ $$\to (x-9)(x-1)=0\to x=9, x=1$$

and we get our wrong solution, because when we square both sides, we must realise that we create a second set of wrong (extraneous) solutions.

How does this relate to your equation? Well, solving, we get:

$$\sqrt{x^2+9}=-1\to x^2+9=1\to x^2=-8\to x=\pm i\sqrt 8$$

Plugging this in we get:

$$\sqrt{-8+9}=-1\to\sqrt1=-1$$

which doesn't hold because of how$\sqrt x$ is defined (you have to take the negative square root)

Answer 2

Actually, it can be true if you choose $\sqrt{z}$ to mean a branch of the complex square-root which, for $z\in \mathbb{R}$, outputs its negative squareroot. In that case, it is even true that $\{\pm \sqrt{8}i\}$ is also a set of solutions.

The weirdness, as Martlund pointed out, comes from the fact that you need to define your square-root function, since $z\mapsto z^2$ is not injective.

Answer 3

Not that $x^2=4 \implies (x-2)(x+2)=0 \implies x=\pm 2$. Similarly $x^2=4 \implies x=(4)^{1/2}\implies x= \pm \sqrt{4} =\pm 2.$ But $\sqrt{4}=+2, \sqrt{1}=+1$. Also $(1)^{1/2}$ has two branches: $\pm 1$, and $(1)^{1/3}$ has 3 branches:$ 1,\omega, \omega^2.$

Answer 4

The square root function $x\mapsto\sqrt x$ is well-defined only for non-negative real values of $x$; and by convention, it always takes a non-negative value. So $\sqrt{x^2+9}$ can't be $-1$.

However, the solutions $x=\pm 3i$ are legitimate, because $x^2+9$ is real and non-negative.