# Why is $\pm x = \sqrt y$ incorrect and $x=\pm \sqrt y$ correct?

Let $$x^2=y$$, then $$x=\pm\sqrt y$$. But why can't it be $$\pm x=\sqrt y$$? I started thinking about this when i encounter this answer to a question but I didn't really understand it. In the answer, it says that there are two possible definitions for the notation $$√$$:

1. For any positive real number $$a$$, $$\sqrt a$$ is defined as the square roots of $$a$$
2. For any positive real number $$a$$, $$\sqrt a$$ is defined as the positive square root of $$a$$

By the first definition, $$\sqrt {16}= \pm 4$$. It also says that if I use the first definition,then I will encounter some problems in the future and that’s why we use the second definition. In the answer it says that solving for $$x$$ in $$x^2 - \pi =0$$ for $$x>0$$, by doing $$x=\sqrt \pi$$, would be incorrect. I still can’t understand why $$x=\sqrt \pi$$ would be incorrect if I use the first definition.

For me, $$\pm x = \sqrt y$$ and $$x=\pm \sqrt y$$ looks like they mean the exact same thing, but I suppose they don’t. Why is that?

• I think the source of confusion here is not the definition of $\sqrt{\cdot}$, but what you mean by $\pm$. $$x = \pm \sqrt{y} \Leftrightarrow x = \sqrt{y} \vee x = -\sqrt{y}$$ but $$\sqrt{y} = \pm x \Leftrightarrow \sqrt{y} = x \vee \sqrt{y} = -x$$ So, the second expression only gives a valid alternative (depending on the sign of $x$) while the firs always gives two alternatives. Commented Nov 3, 2021 at 9:17
• $(1)$ the square root is the non-negative solution by convention $(2)$ $\pm x=\sqrt{y}$ is the same as $x=\pm \sqrt{y}$ , but it is stange to have $\pm x$ on the left side if we want to have the value of $x$. $(3)$ If $x>0$ , then we only have one solution which is the square root without a sign. Commented Nov 3, 2021 at 9:17
• @Peter I think your (2) is the true answer here: It's correct but strange. Why not make an answer? Commented Nov 3, 2021 at 9:49
• Why so much downvotes on the question (-3) ? It is in mathjax, OP explains thoroughly his source of confusion and ask for clarifications, this is fine for me.
– zwim
Commented Nov 3, 2021 at 10:34
• @PierreCarre so from what I understand, $\pm x = \sqrt y$ is pretty much the same thing as $x = \pm \sqrt y$? Commented Nov 3, 2021 at 14:11

Neither is incorrect. $$x =±\sqrt{y}$$ is the same as $$\sqrt{y}=±x$$.

However, you'd usually want to have the variable on one side (usually left) and its solution(s) on the other side (usually right).

Thus, one would normally interpret $$x=±\sqrt{y}$$ as $$x$$ is the variable and $$±\sqrt{y}$$ are its solutions while $$\sqrt{y} = ±x$$ would be interpreted as $$\sqrt{y}$$ is the variable and its solution is $$±x$$. Both are of course, equivalent ways of saying the same thing.

• so, it’s just a convention? saying $\pm x = \sqrt y$, for $x ≥ 0$, wouldn’t be wrong? Commented Nov 3, 2021 at 14:09
• @Mohammadmuazzamali Why $x≥0$? Did you mean $y≥0$ instead? Because $x$ can be any real number and only $y$ needs to be non-negative (for $\sqrt{y}$ to be defined) in the equation $x^2 =y \Leftrightarrow x = ± \sqrt{y} \Leftrightarrow ±x = \sqrt{y}$. In that case, yes. It's a convention to want to have the unknown on one side and its solutions on the other. Commented Nov 3, 2021 at 23:57

Let $$y\geq0$$. If $$\sqrt{y}$$ is defined as the set of solutions to the equation $$y=x^2$$ and $$x$$ is one solution then I would say that $$\sqrt{y}=\pm x$$ is perfectly sensible notation (since $$x$$ and $$-x$$ are the only solutions). More commonly, however, $$\sqrt{y}$$ is defined as the unique non-negative solution to $$y=x^2$$. In this case, if $$x^2=y$$ then $$x=\sqrt{y}$$ or $$x=-\sqrt{y}$$. In my opinion one should avoid writing either of the expressions "$$x=\pm\sqrt{y}$$" and "$$\sqrt{y}=\pm x$$". My point: $$x$$ is some number, but what is $$\pm\sqrt{y}$$, and similarly, $$\sqrt{y}$$ is some number but what is $$\pm x$$?

$$\sqrt {x}$$ is the "principal root." That means that $$\sqrt x \ge 0.$$ (Or, at least it is until you learn about complex numbers.)

$$x^2 = 4$$ is multivalued and has two solutions. That is $$x = \pm \sqrt 4 = \pm 2$$

When you write $$\pm x = \sqrt y$$ this requires that it be possible that $$\sqrt y$$ be negative, which it can't.

$$x = \pm \sqrt y$$ says $$\sqrt y$$ is always positive, but $$x$$ can be negative.

• "When you write $±x = \sqrt{y}$ this requires that it be possible that $\sqrt{y}$ be negative, which it can't." I don't think so. Writing $\sqrt{y} = ±x$ means $\sqrt{y}$ is "$x$ or $-x$", not "$x$ and $-x$" so it doesn't require anything of the sort. Commented Nov 3, 2021 at 9:40
• @Ibrahim When we say $x = \pm 2$ then that means that both $2,$ and $-2$ are acceptable values for $x.$ Commented Nov 3, 2021 at 9:44
• Not quite, the sign $±$ means "plus or minus" where or is inclusive. So $x=±2$ simply means $x=2$ or $x=-2$ or both. Commented Nov 3, 2021 at 9:52

There has been a lot of nonsense posted here. The facts are these:

The notation

$$a=\pm b$$ is simply a universally accepted shorthand for $$a=b\text{ or }a=-b$$

The notation $$\pm a=b$$ has no universally recognised meaning, and should never be used.

• Can $\pm a = b$ be understand as $a = b$ or $-a = b$? that also seems pretty reasonable. Commented Nov 4, 2021 at 7:05
• @Mohammadmuazzamali: There is no need for it, and it is not normal, or expected. Also, if we allow $\pm a=b$, then we should also allow $c=\pm a=b$; but what would that mean? $c=b$, or $c=\pm b$? Better not to use it $-$ I can't see why you would want to. Commented Nov 4, 2021 at 11:17