# square root and sign

My question is very very basic but, for the life of me, I'm confused for whatever reason.

I know that if $$x^2 = 5$$ then +$$\sqrt{5}$$ and $$-\sqrt{5}$$ are the solutions for $$x$$.

The reason, as I understand it, is that, in a function, $$x$$ can have two values while $$y$$ can only have one. And both values, when squared equal to $$5$$.

I also know that $$\sqrt{4} = 2$$ and $$2$$ only. The reason is we deal now with a square root function which only leads to positive values on the $$y$$ axis, otherwise, it wouldn't be a function as $$y$$ would have two values as in $$\sqrt{x} = +y$$ and $$-y$$, which isn't possible.

Now what about this then?

$$y= \sqrt{x^2}$$ ?

If I consider, for the sake of the example, that $$x^2 = 4$$ then we've already said that $$\sqrt{4}$$ only equals $$2$$, not $$-2$$. So the answer should be $$x$$ only, not both $$x$$ and $$-x$$. This makes sense somehow otherwise I would get two values for $$y$$ which is forbidden when it comes to functions.

So this leans towards the fact that $$y= \sqrt{x^2}$$ is definitely equal to $$x$$.

But then I see this on a youtube course:

And this totally lost me, even though it's trivial.

I don't know why that second line is using the absolute value (probably to emphasize that it remains a positive $$x$$ which is consistent with what I concluded here above) but then the last line considers $$-x$$ as an eligible value this time and the part is beyond me. I know I'm overthinking it but I lost confidence with square roots right now. I need to get back to the basics.

Thanks for your patience and input.

• When we take square roots, we really mean the so-called principal square root, when the result is positive. In general, $\sqrt{x} = |x|.$ May 8, 2020 at 3:47
• Ok thank you very much but then, how come it turns out as a -x on the third line of this limit, all of a sudden ? May 8, 2020 at 3:55
• Because $x\to-\infty$ says that we're looking at negative numbers $x$. May 8, 2020 at 4:10

## 1 Answer

We say $$\sqrt{x}$$ refers to the principal root of $$x$$, which in the case of the positive real numbers refers to the root of $$x$$ that is greater than zero. So, $$\sqrt{x^2}$$ is the principal root of $$x^2$$. Now as you know, the square roots of $$x^2$$ are $$x$$ and $$-x$$, but only one of those is positive! Since the principal square root only cares about the positive root, it is the same as the absolute value in this instance.

• Ok thank you very much but then, how come it turns out as a -x on the third line of this limit, all of a sudden ? May 8, 2020 at 3:54
• @BachirMessaouri I guess what's happening here is that $|x|=-x$. Remember this happens whenever $x \le 0$. Since the limit is approaching negative infinity, $x \le 0$. May 8, 2020 at 3:55
• That's the part that is confusing me a lot. For 2 reasons. first because this means the square root of x^2 could lead to -x because it is an option eventually (proof on the third line), even though the scare root should prevent that. And second, I suspected it's because the limit leans towards - infinity but it should mean the value of x is negative not that the sign of the variable x is negative. Of course you are right, no question about it, I just fail to understand the logic. May 8, 2020 at 4:00
• Ok, got it after a few minutes of thinking ;-) Thank you. May 8, 2020 at 4:14