GRE Question ($|x|$ vs $\sqrt{x^2}$)

Question

I just finished taking a Kaplan GRE Practice Test, and I encountered an interesting question with a very vague solution.

If you have taken the GRE, then you are probably familiar with the following question format. The question shows two quantities labeled as A and B, and you need to choose whether A is greater, B is greater, they are the same, or there is not enough information.

The question was the following

Quantity A: |x| Quantity B: $\sqrt{x^2}$

I answered that there is not enough information because the square root of $x$ has two solutions, the negative and positive root, whereas $|x|$ is always positive. Kaplan said that the solution is that they're the same (it looks like they are assuming that the only solution to $\sqrt{x^2}$ is the principal root.

Who is right and why?

Thanks!

Surely B is $\sqrt{x^2}$, no? Otherwise, for $x = 4$, the two are evidently different. In general, $\sqrt{x}$ denotes the nonnegative square root; if $x$ is negative, then it's undefined (at least until you enlarge the domain of discourse to include complex numbers). — John Hughes, Jun 29 '15 at 01:26
It's a convention. A positive number has two square roots, but nonetheless if $x>0$ then $\sqrt x$ denotes "the" square root, namely the positive one. — David C. Ullrich, Jun 29 '15 at 01:35
Re: "it looks like they are assuming that the only solution to $\sqrt{x^2}$ is the principal root." --
There is no 'solution' because it is not an equation! When $a \ge 0$ the notation $\sqrt{a}$ refers to a quantity, namely the non-negative root $y$ of the equation $y^2=a$. The other solution happens to be $-\sqrt{a}$. — Clive Newstead, Jun 29 '15 at 02:02

Moya · Accepted Answer · 2015-06-29T01:45:06.653

7

It is most likely a mistyping on the exam's part or you, as I believe it should be $|x|$ vs. $\sqrt{x^2}$, as your statement doesn't work for $x=2$, as $|2|=2$ whereas $\sqrt{2}<2$.

In this case the quantities are equivalent, which you can see by always taking the positive solution to the square root, it must be that $\sqrt{x^2}=x$ if $x\geq 0$ and $\sqrt{x^2}=-x$ if $x<0$, which exactly defines $|x|$.

EDIT: For a more thorough explanation, let $y=\sqrt{x^2}$. Then clearly $x^2-y^2=0$. By expanding we get $(x-y)(x+y)=0$, so that either $y=x$ or $y=-x$, and we always take whichever one is positive.

EDIT 2: Here is a web cache from ETS's website (which writes the GRE), which states that, in their context, the symbol $\sqrt{\;\;}$ always is positive.

edited Jun 29 '15 at 01:45

answered Jun 29 '15 at 01:28

Moya

5,248

Yes, but why is it the case that you always have to take the positive solution; why not the negative solution? – Vincent Jun 29 '15 at 01:35
As @David C. Ulrich mentions, it is just a convention that we usually take the square root of a number to be positive as this is the "principle" square root. I have added a link from ETS which states as such. – Moya Jun 29 '15 at 01:41
3

Simple.
$\sqrt{x}$ is always positive. When in an equation, you encounter $y^2=x$, then you'd write that as: $y = + $or $ - \sqrt{x}$. If you notice, the solution for the equation itself mentions + or - as addenda to the $\sqrt{x}$; the $\sqrt{x}$ itself is always the positive root of x by convention (the function is defined like that).
– Panglossian Oporopolist Jun 29 '15 at 01:43

symplectomorphic · Answer 2 · 2015-06-29T19:21:54.820

You are confused when you write they are "assuming that the only solution to $\sqrt{x^2}$ is the principal root." It makes no sense to speak of a "solution" here, because you have no equation.

By convention, the symbol $\sqrt{x}$ always denotes the nonnegative square root of $x$. We make that convention, among other reasons, so that $\sqrt{x}$ defines a function. In your problem, the symbol $\sqrt{x^2}$ stands for the nonnegative square root of $x^2$. So we must take the absolute value of $x$ to ensure the result is nonnegative. In short, the equation $\sqrt{x^2}=|x|$ is an identity. It holds for all real values of $x$.

(By contrast, the equation $\sqrt{x^2}=x$ is not an identity. It fails to be true for negative numbers. For example, $\sqrt{(-5)^2}\neq-5$. Again, that is because $\sqrt{(-5)^2}$ denotes the nonnegative square root of $(-5)^2=25$, which is 5.)

Many people abuse this notation. For example, many people write $\sqrt{4}=\pm2$. Such an equation is not only wrong but meaningless. $\sqrt{4}$ denotes a unique number. And in any case, no number can equal both $2$ and $-2$.

If you want to use the word solution in this context, you can say the following. The equation $x^2=c$, where $c$ is a positive constant, has two solutions, namely $\sqrt{c}$ and $-\sqrt{c}$. Regardless, $\sqrt{c}$ denotes a unique number.

GRE Question ($|x|$ vs $\sqrt{x^2}$)

2 Answers2