Is $\sqrt{x^2}$ always $\pm x?$

Question

I am wondering if this holds in every single case: $$\sqrt{x^2} = \pm x$$

Specifically in this case: $$\sqrt{\left(\frac{1}{4}\right)^2}$$ In this one we know that the number is positive before squaring, so after removing the square and root shouldn't we just have: $$\frac{1}{4}$$

Also in a case such as this: $$\sqrt{(\sqrt{576})-8}$$ we have $$\sqrt{\pm24-8}$$ which is either $\sqrt{16}$ or $\sqrt{-32}$ but since we care about real, principal roots only, can't we say that $\sqrt{16}$ is actually $4$ but not $-4.$

Am I mistaken somewhere in my reasoning?

Thanks!

Answer 1

The principal square root of $x$ denoted by $\sqrt x$ is defined to be the non-negative square root of $x$. Thus $\sqrt{25}=5$, $\sqrt{3^2}=3$ and $\sqrt{(-3)^2}=\sqrt 9=3$.

Note that, $\sqrt {x^2}$ can be $x$ or $-x$ depending on whether $x$ is positive or negative or zero. For example, $\sqrt{7^2}=7$ and $\sqrt{(-7)^2}=7$. It should be clear that when $x\geq0$, $\sqrt {x^2}=x$, but when $x<0$, $\sqrt {x^2}=-x$. This can be briefly written as $\sqrt{x^2}=|x|$.

When you're talking about finding all the square roots of a positive number, then indeed there are two solutions. In other words, you're looking to solve $x^2=a$ for some positive constant $a$. Then $x$ can either equal $a$ or $-a$. This is usually written as $x=\pm a$. Note that for $a=0$, there is only one solution and for $a<0$ there are none in $\mathbb R$.

Answer 2

I assume that in your question, you meant to ask about $\sqrt{x^2}$ rather than $\sqrt{x}$. My answer is based on that assumption.

It is inaccurate to say (in the case that you are interested only in real, principal roots) that $\sqrt{x^2}=\pm x$; the correct thing to say is that $\sqrt{x^2}=\lvert x\rvert$.

This is a common confusion; I think that it usually results from the fact that if you have $\sqrt{x^2}=y$, then you get $x^2=y^2$... and this has two different solutions: $x=y$ and $x=-y$.

That is not the same thing as saying $\sqrt{(-3)^2}=\pm 3$; clearly, if you care about principal real roots only, then $\sqrt{(-3)^2}=\sqrt{9}=3$. Note, however, that $3=\lvert-3\rvert$, though; this is why you cannot simply say that $\sqrt{x^2}=x$.

Answer 3

As a matter of fact, the notation $\sqrt a$, where $a \in \mathbb R$ is non-negative, is defined to be the unique non-negative $b \in \mathbb R$ such that $b^2 = a$.

So $\sqrt {x^2}$ is either $+x$ (if $x \ge 0$) or $-x$ (if $x < 0$).

With this clarification, you can see that your equation

$$\sqrt{\sqrt{576}-8} = \sqrt{\pm24 -8}$$

is wrong (if it means anything at all). It should be simply:

$$\sqrt{\sqrt{576}-8} = \sqrt{24 -8} = 4$$