Absolute Value Basic

Algebra Level 2

Let x x be a real number. Which of the answers is equal to the below?

x |-x|

None of the above x 2 \sqrt{x^2} x -x x x ± x \pm x

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

7 solutions

Bufang Liang
Aug 28, 2015

The initial misleading answer is x x . If x x is a negative number, then

x = x |-x| = x

is false. The left side is positive while the right side is negative.

However, the answer isn't ± x \pm x either, as it wouldn't specify what cases it is positive and what cases it is negative. This answer is ambiguous so it is not correct.

The correct answer is x 2 \sqrt{x^2} because it forces the value to be positive regardless of whether x x is positive or negative. Alternatively, this can also be written as x |x| .

The most definitive way to visualize this is probably to just look at the two graphs of x |-x| and x x . If they were in fact equal, you should get the exact same graph, but you don't.


EDIT: It appears there's a massive amount of reports on my problem for some reason. Let me help explain this in more detail to address the most common complaints.


Complaint 1: x 2 \sqrt{x^2} is actually x x , or also x -x .

Response: There is a very fundamental difference between saying "there are two numbers we can square to get 25" (solutions of x 2 = 25 x^2 = 25 ) and "what is the square root of 25?" ( 25 = x \sqrt{25} = x )

Any equation for which you solve for solutions, you are finding all of the values you can plug in in order to make the equation true . You can plug in both 5 5 and 5 -5 for the first equation, and there is no ambiguity that both work. However, in the second equation, in order to compare the left and right sides, there needs to be no ambiguity as to what 25 \sqrt{25} is. How would you evaluate ± 5 = 5 \pm 5 = 5 ? Is this equation true or not? This is one of the reasons why we define functions, to remove that ambiguity. It just so happens we define the square root function to take the positive value, so there is only one solution for the second equation (which is 5 5 ).

Davy Ker below me explained this quite well: "...what would be the use of a symbol if every time you use it, it could represent one of two numbers?"

After reading the response to complaint 3 , one might realize the fundamental difference between the meaning of these two equations:

x 2 = 25 x^2 = 25 asks for all of the points that are a distance of 5 from the origin. (which are both 5 and -5)

25 = x \sqrt{25} = x asks for the distance the number 5 is from the origin. (which must be 5)


Complaint 2: There's no difference between ± x \pm x and x 2 \sqrt{x^2} . Both are correct.

Response: In mathematics, we use ± \pm to denote that both solutions are correct. In this problem, only one of the two solutions can ever be right for any particular value of x x , so this can't be right. On top of that, this answer is ambiguous like I mentioned above, and there are strictly better answers that are not ambiguous.

On top of that, if x 2 = ± x \sqrt{x^2} = \pm x , then there would be two correct answers, which is not possible. I actually deleted the problem the first time around to add ± x \pm x as an answer choice in order to deter people from thinking those two were equivalent.


Complaint 3: It should just be the "positive value" of x x .

Response: The whole point of this problem is to test whether people actually understand what absolute value is, and demonstrates the importance of technical definitions and concepts.

Absolute value is defined as a number's distance from zero, or magnitude. It is not simply taking the positive value. In this problem, x x is a real number, but what if x = 1 + i x = 1+i ? How would you understand absolute value here? If you treated complex numbers or vectors with the same idea of just "taking the positive value," not only would you be confused but you might also get wrong results.

Absolute value appears in many places in mathematics, and is used on a pretty much daily basis in physics and physics calculations. For example, when you add up forces, you must add the vectors together. By considering the magnitudes of the forces, you can find the range of magnitudes of the sum force. Without absolute value, it's very difficult to do even basic physics.

This also confirms that x 2 \sqrt{x^2} is the right answer, as it is the result of taking a 1-dimensional distance. To show why, imagine we are looking for the distance between the points ( 0 , 0 ) (0,0) and ( x , 0 ) (x,0) . This is equivalent to the distance between 0 0 and x x on the number line. Plugging into either the distance formula or Pythagorean's theorem (you should be using Pythagorean) you get:

( x 0 ) 2 + ( 0 0 ) 2 = x 2 + 0 2 = x 2 \sqrt{(x-0)^2 + (0-0)^2} = \sqrt{x^2+0^2} = \sqrt{x^2}

Moderator note:

Another way to verify the answer is to test out various values of x x .

For example, if x = 1 x = 1 then x = 1 |-x| = 1 , and so x -x and ± x \pm x are not valid answers.
If x = 1 x = -1 , then x = 1 | - x | = 1 and so x x is not a valid answer.

This leaves x 2 \sqrt{ x^2 } as the only valid answer. We verify that this is true because

x 2 = x = x , \sqrt{x^2} = |x| = |-x|,

where the equality is as real numbers and not as variables.

The square root of a number can be negative. sqrt(x^2) is both x and -x. We often throw away the negative root, but both positive and negative values are correct. Therefor, the answer should be "None of the above".

Stacey Greenstein - 5 years, 9 months ago

Log in to reply

I disagree with you:

\sqrt{} : R + \mathbb{R}^{+} \rightarrow R + \mathbb{R}^{+}

There's a difference between the function and the concept of a square root. Clearly, the answer x ² \sqrt{x²} is the function, which forces the result to be a positive number.

Valentin Michelet - 5 years, 9 months ago

Actually, the √ sign denotes the principal root, the positive square root of a number. What would be the use of '√' if every time we used it we couldn't be sure what we're talking about?

Davy Ker - 5 years, 9 months ago

I tend to agree with you stacey. Worst case, I would expect the answer to be x and it is definitely not what was mentioned.

Sundar Bala - 5 years, 9 months ago

This is why I have come to hate Brilliant. These are tricksy, not mathy.

Max Harris - 5 years, 9 months ago

Log in to reply

In a way, I understand how you feel. Math should be about testing your ability to logically think, not to trick people with unfair problems. However, this problem is not one of those, even though it may seem like it.

When you do just about any math problem, you always need to be aware of particular outside cases and check your work. This problem is no exception. Any of the wrong answers you could've eliminated by simply plugging in a number for x x , so there's really no excuse.

What makes this even more fair is that the entire question is focused on this concept. I didn't "hide" the expression in a larger question or anything, I removed everything else and exposed it so it is the only thing to pay attention to. People who are "trigger-happy" or don't understand absolute value will get the problem wrong, which should hopefully teach them that in math, you need to slow down to approach a question logically and trust that logic, not gut reaction. (And also, if you don't understand what absolute value is, figure it out before doing the problem or read up on why you got it wrong)

The only other way where I can see this might be even remotely unfair is if someone didn't know that x 2 = x ± x \sqrt{x^2} = |x| \neq \pm x . However, if x 2 = ± x \sqrt{x^2} = \pm x , then there would be two correct answers, which is impossible. I actually deleted this problem the first time around in order to add ± x \pm x as an answer choice to help prevent this possible unfairness.

This is a topic and idea I care deeply about, hence the detailed response. The reason why I chose to write this problem in the first place is to expose the poor logical thinking or appreciation of rigor in most people. (This problem got reported almost 30 times in one day, all of which were resolved as there being nothing wrong with this problem!) Even something as basic as absolute value, which you probably learned in elementary or middle school, you can fundamentally misunderstand or get too confident and slip.

Bufang Liang - 5 years, 9 months ago

Log in to reply

Simply put, this is about teaching principals and the reasoning behind them. But perhaps to seem less tricksy, the question should have been worded a bit differently... because "Let be a real number. Which of the answers is equal to the below? |-x| well... in any real situation |-x| = x I applaud your attempt to drive the idea behind absolute value home though, it's something I had forgotten how to properly calculate because of the ease of simply remembering the absolute value of a number is that number in a positive form. Lots of criticism here, lets try examining why some of us have the need to correct, and not to listen? "A wise man can entertain an idea, without having to accept it" this is the only true way to humble yourself to learn something different what you are certain you know. That's why I stepped out of Mathematics, the general attitude of (especially self proclaimed) Mathematicians is "superiority" not "Grace" <--BTW This guy is top 1% in the USA in Math skills according to the California Achievement Test.

Chris Lesmerises - 5 years, 9 months ago

Log in to reply

@Chris Lesmerises "well... in any real situation |-x| = x"

Fundamentally untrue.

If x < 0, |-x| = -x

Matt Steele - 5 years, 9 months ago

The answer is x. |-3| is 3 correct. Your answer adds a level of unnecessary complexity as the answer as the square root of 3 squared is 3.

Mary Beth Stephens Wilson - 5 years, 9 months ago

Log in to reply

Incorrect. Consider x = 3 ( 3 ) 3 x= -3\Rightarrow \lvert -(-3) \rvert \neq -3 .

Anders Therkelsen - 5 years, 6 months ago

But the the range of the modulus function is always positive? It is like mod(-x)=mod(x) where mod(x)=x.

Vibodh Jadhav - 5 years, 9 months ago

Log in to reply

Mod(x) =x only if x>=0, if x<0, mod(x) =-x.

Hely Sousa - 5 years, 9 months ago

The, what if x=1+i, argument is invalid because the question states x is real and in the real plain x=(x^2)^1/2. There is no special case where the left and right side aren't equivalent. So x is a valid answer.

Herman Pittman - 5 years, 9 months ago

Log in to reply

If x<0 your answer is wrong.

Hely Sousa - 5 years, 9 months ago

This is the best explanation. It makes you think about the problem in a left side/right side way which makes it clear why (x^2)^1/2 is the only answer.

Herman Pittman - 5 years, 9 months ago

Very good explanation, thank you.

Ryan Holmes - 5 years, 7 months ago

point of writing pow 2 and square are hidden, unless you remember those rules..

Zaf Haidari - 5 years, 7 months ago

Baloney - The question not NOT about solving an equality.

Pierre Fiorini - 5 years, 9 months ago

Log in to reply

"Which of the following is equal to |-x|?" That sounds like we're solving an equality to me...

Davy Ker - 5 years, 9 months ago

That is not right mathematically, the value either in bracketed or square rooted will neglect the negativity. Here x is the same than the root of x^2 Regardless of the sign of x because you asked about the equivalence not about the nature of x itself.

Ibrahim Ghalib - 5 years, 9 months ago

I'm sorry but I failed to comprehend your rationale behind your dispute of complaint number 3. Absolute value of any integer positive or negative is always positive, correct? Yes, you can complicate it whenever the applications arise but fundamentally it is always correct. So you're right that in the case of a negative number for x, the |-x|=x answer is incorrect but I do think the answer should be just +x.

Drew Davis - 5 years, 9 months ago

Log in to reply

The reason why we care about application is because the whole point of mathematics is to assist problem solving in the real world. Almost everything interact with on a daily basis, including your car, your house, your plumbing and electricity, your food, your computer, your phone, all of these things are a result of scientific progress, which is not possible without applications in math. This is why we care, and this is why we need to understand definitions conceptually and technically.

But what is even more baffling to me is why you still think the answer is x x even though you yourself just admitted that it isn't right:

"...in the case of a negative number for x, the |-x|=x answer is incorrect"

"...I do think the answer should be just +x."

You just contradicted yourself in one sentence, so I don't really know how else to respond to that.

Bufang Liang - 5 years, 9 months ago

Log in to reply

In other words, what I meant to say was have x always positive. x and +x are not the same so there is no contradiction. I understand there are applications, but again, this is fundamentals. Let's not deter from the fact that absolute value will always result in a positive number. Right? So positive x on the other side would make this equation true, no? Have I made that a little clearer if possible? I'm just saying there's another solution besides taking the square root of a square.

Drew Davis - 5 years, 9 months ago

Let's say that x=-2. In solving |-x| = |-(-2)| = |2| = 2. If x is -2, you're looking at either -x or |x|.

Ronald Sweeney - 5 years, 7 months ago

root(square(x)) is nothing but + or - x. root(25) can be + 5 or -5.

Karthikeya Kaluvakolanu - 5 years, 9 months ago

Good work .Upvoted.

Sai Ram - 5 years, 8 months ago

Yes but square root of a number has plus and minus as the answers. So the suggested answer is incorrect! All other options were not valid other than non of these.

Sarge Sandhu - 5 years, 9 months ago
Davy Ker
Sep 2, 2015

x 2 \sqrt{x^2} is the only option to ensure our answer is positive, which the modulus requires. x 2 x^2 is always positive and √ always gives the positive square root: x x if x 0 x≥0 , x -x if x < 0 x<0 ,

x 2 \sqrt{x^2} denotes the positive square root of x 2 x^2 . It's a common misconception that a \sqrt{a} can be both the positive or negative root of a non-negative a, but what would be the use of a symbol if every time you use it, it could represent one of two numbers? This more precise definition allows you to specify the negative square root with " a -\sqrt{a} "

Wikipedia on square roots

to me if you want the magnitude then abs(-x) = sqrt((-x)^2) which gives you a positive x. The answer you give ignores the (-)

Rick Mercado - 5 years, 9 months ago

Log in to reply

( x ) 2 = x 2 (-x)^2 = x^2

Davy Ker - 5 years, 9 months ago
Rishabh Mishra
Aug 30, 2015

for x>0, number inside mod is negative, so the result will be positive.

for x>=0 number inside mod is positive, so the result will be positive.

in both the cases value of x is positive. So we conclude that result is |x| (or sqrt(x^2)), as sqrt(x^2) is also |x|.

Leslie Koller
Sep 3, 2015

The radical symbol when written is not read "square root", rather "principle square root", which is ALWAYS positive. The problem's author is correct in every statement.

Brian Lee
Sep 2, 2015

If x is negative, you need to square it and square root it (as the value is positive).

Matt Steele
Sep 2, 2015

It's clear that most people on this thread do not understand how radical signs work. A radical with an even index ALWAYS returns the positive root. Period. That's how the symbol is used. Consider what happens when you use a graphing calculator to graph y =sqrt (x). If the radical sign allowed for two results, the resulting graph would be a parabola opening to the right. But that isn't what results. Instead, you get only the top half; in other words, the POSITIVE results.

but this question reveals the principal of WHY.

Chris Lesmerises - 5 years, 9 months ago

Log in to reply

Not really. It relies on the definition of a radical sign. It doesn't provide any rationale whatsoever.

Matt Steele - 5 years, 9 months ago

Log in to reply

I address this in the complaints section and include a handful of examples that confirm it.

"On top of that, if x 2 = ± x \sqrt{x^2} = \pm x , then there would be two correct answers, which is not possible. I actually deleted the problem the first time around to add ± x \pm x as an answer choice in order to deter people from thinking those two were equivalent."

Though I will admit it is not necessarily deduced from the mathematics itself, and does require a bit of extra reasoning.

Bufang Liang - 5 years, 9 months ago
Hadia Qadir
Aug 30, 2015

Function above is 'if you put - you'll get +, and if you put + you'll get +.

And the answer is equivalent to sqrt(x^2)

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...