Confusing square roots

Algebra Level 2

$\begin{aligned} \text{Statement 1: } \sqrt{x^2} &=& x \\\\ \text{Statement 2: } \sqrt{x^4} &=& x^2 \\\\ \text{Statement 3: } \sqrt{x^6} &=& x^3 \\\\ \text{Statement 4: } \sqrt{x^8} &=& x^4 \\\\ \end{aligned}$

If $x$ is a real number, which of the statements above are true?

Statement 1 and Statement 3 Statement 1 and Statement 4 Statement 2 and Statement 4 All the statements are correct

8 solutions

Anuj Yadav
Nov 2, 2015

Since √x^2 = |x| so In first statement |x|=x which is not true. Second statement can be written as √(x^2)^2 , this equals to |x^2|=x^2 which is always true. Similarly Third statement is |x^3|=x^3 which is not true in case x = -1. Fourth statement |x^4|=x^4 which is true. Hence 2 & 4 is true

I think your understanding and explanation are correct but, when you use a variable it could be negative or positive but that's "in" the variable, so x could be 1 or -1 (for example) but the minus sign is IN the variable you don't know the sign until you substitute the value for the variable, other reason that makes me believe that the answer is wrong is that you could express the roots as powers, and then we get:

(x^2)^(1/2) as we know a power of another power its multiplied so then we have 2*1/2=1 so (x^2)^(1/2) = x And the same for the other statements:
(x^4)^(1/2) = x^2
(x^6)^(1/2) = x^3
(x^8)^(1/2) = x^4

So the right answer should be "All the statements are correct"

Isaac Salinas - 5 years, 7 months ago

I am agree with Isaac. all statements are correct since the condition was not given that whether it is true for integers

Ashwin Chavan - 5 years, 7 months ago

@Isaac Salinas Actually, the square root function has it's range $\ge$ 0, therefore, the result has to be positive. Surely, the sign is "in" the variable itself, and thus $x$ may be positive or negative, and that itself is the explanation to your query, as the square root of $\sqrt { { x }^{ 2 } } =\left| x \right|$ , (as it has to be positive, thus the absolute value). Thus, $\sqrt { { x }^{ 6 } } =\left| { x }^{ 3 } \right| \begin{cases} ={ x }^{ 3 }\quad if\quad x\ge 0 \\ =-{ x }^{ 3 }\quad if\quad x<0 \end{cases}\quad as\quad { x }^{ 3 }\ngeq \quad 0\quad \forall \quad x\in \Re \\ \sqrt { { x }^{ 8 } } =\left| { x }^{ 4 } \right| ={ x }^{ 4 }as\quad { x }^{ 4 }\ge 0\quad \forall \quad x\in \Re \quad \quad \quad \quad$

Abhijeet Verma - 5 years, 7 months ago

Nicely done!

Pamela Barnes - 5 years, 7 months ago

@Isaac Salinas No. All statements are not true. It's very important to be clear on that. @Abhijeet Verma has explained it here.

Harish Sasikumar - 5 years, 6 months ago

The symbol implies principal root unless stated otherwise.

Phoebe Liu - 5 years, 7 months ago

Your reasoning is incorrect, when you take the square root of x^2 you get +/-2 because you can square both 2 and -2 and get the same answer of 2. The same thing is true for the absolute value function. taking the square root of odd powers only produces one answer because you can't get the same result by plugging in positive and negative numbers

Calvin Francis - 5 years, 7 months ago

See anuj yadav's explanation above.

Pamela Barnes - 5 years, 7 months ago

It is still wrong.

Nicholas Findlay - 5 years, 7 months ago

@Nicholas Findlay – The square function is defined as $\sqrt{x^2}=|x|$ ,

Abdur Rehman Zahid - 5 years, 7 months ago

@Abdur Rehman Zahid – √ is a special function that returns the principal square root . The square root of x^2 is ±x, but √x^2 is x. This is not very complicated... https://en.wikipedia.org/wiki/Square_root

Paul Alberti-Strait - 5 years, 7 months ago

@Paul Alberti-Strait – Actually, your link repeats what has already been stated, contrary to what you have written.

$\sqrt{x^{2}}=|x|$

NOT $\pm x$ as others continue to incorrectly write, and is potentially confusing. You obviously understand principal square root, but others on this thread continue to struggle. Peace!

Pamela Barnes - 5 years, 7 months ago

@Pamela Barnes – I don't think that what I wrote is contrary to the link, but I can agree that it might be confusing as written. The issue is that every positive number has two square roots, one of which is the principal root, which is positive and which is identified by the √ sign. Because √(x^2) = |x|, x = ±√(x^2). Importantly, the ± sign belongs in front of the √ sign, since √(a) only returns the principal square root of a. So, I agree that √(x^2) =/= ±x , but there isn't anything inherently confusing about the ± sign in explaining why this is true.

Paul Alberti-Strait - 5 years, 7 months ago

@Paul Alberti-Strait – It's very confusing if you do not include the conditions.

$\sqrt{x^{2}}=|x|$

which is

$x$ , if $x \geq 0$ ,

or $-x$ , if $x<0$

It is mathematically incorrect to write it otherwise.

I understand that you are referring to solutions of mathematical equations involving two solutions when square root is involved. In the context of this particular problem, however, it is incorrect. $\sqrt{x^{2}}$ has only one positive "solution" in this problem, which must be written as $|x|$

Sorry to get so pedantic, but there's been a lot of misunderstanding of symbolism in this thread.

Pamela Barnes - 5 years, 7 months ago

@Pamela Barnes – I guess I just don't see how any of that is inconsistent with anything I wrote. I'm also not sure why conditions need to be specified; what information does it add that isn't contained in the absolute value statement? The entire reason to use the +/- sign is to show how it is different from the scenario in this question. As such, it is a help, not a hindrance.

The statement x = ±√(x^2) logically entails √(x^2) = |x|.

Paul Alberti-Strait - 5 years, 7 months ago

Agree with you

Mohammad Moeen - 5 years, 7 months ago

You are right. I exactly thing the same. (4)^1/2 is not plus minus 2 but only plus 2. Square root define absolute value, so no place of doubt that all statements are correct

tanay gaurav - 5 years, 7 months ago

I totally agree! I can even give an example for statements 1 & 3:

Statement 1: 3*3=9 9/3=3

Statement 3: 3 3 3 3 3 3=729 729/(3 3 3)=729/27 729/27=27=3 3*3

Overall, I think that your formula sumarises the answer well, I left a like. :)

Henry Shorland - 5 years, 6 months ago

Just finding an example that is correct for statements 1 and 3 does not prove that statements 1 and 3 are true for any real number. If x is -1, statements 1 and 3 are not correct.

Paul Alberti-Strait - 5 years, 6 months ago

@Paul Alberti-Strait – Oh, yes, sorry, you're certainly right. I forgot about the negatives. 😎

Henry Shorland - 5 years, 5 months ago

Hey when x^2=y, then x=(+- sqrt y). And here the given case is of sqrt(y). So why do i consider modulus.

Shreyansh Ajmera - 5 years, 7 months ago

Your reasoning is right, but in most other problems on this site, the square root function refers to the principle root. How are we supposed to know that that norm doesn't apply to this question? Standards are necessary bc we aren't psychic.

Paul Alberti-Strait - 5 years, 7 months ago

That norm does apply to this question.Infact it isn't a norm at all,because $\sqrt{x^2}$ is defined to be $|x|$

Abdur Rehman Zahid - 5 years, 7 months ago

You aren't familiar with the definition of a principal square root? https://www.mathsisfun.com/algebra/square-root.html

Paul Alberti-Strait - 5 years, 7 months ago

I see, my reasoning was confused -- but I don't understand why you disputed this point, in that a definition is a kind of norm. In any event, I think you would have an easier time persuading people by using a clear counter-example, e.g., in statement 1, if x = -1, √(1) =/= -1.

Paul Alberti-Strait - 5 years, 7 months ago

What is the difference between Modulo x and x ?

Syed Baqir - 5 years, 7 months ago

x and module x are the same when x >= 0, but when x <0, modulo x will be positive.

Tony Flury - 5 years, 7 months ago

√(x^2) = +-x. √(-2^2)=√4; (-2*-2)=4. Then, first statement isn't true, but not for the fact that √x^2 =|x|, but for the fact that there are 2 possible answers to the square root of a number. In that case, none of the answers are valid, since x^2 and x^4 is possitive.

Rafael Acosta - 5 years, 7 months ago

$\sqrt{x^2}$ = |x| always, because of the rule of principal square root (always positive). It is a common algebra mistake to write that it equals $\pm x$ , it does not. Many teachers don't explain the sequence of solving correctly. I'll do it here for you.

Solve

$x^2 = 9$

$\sqrt{x^2} = \sqrt{9}$

|x| = 3

x = $\pm 3$

The plus-or-minus comes from the absolute value, not from the square root, and appears before the number, not before the x. Make sense?

Pamela Barnes - 5 years, 7 months ago

The plus-or-minus doesn't come from the absolute value, but for the fact that if you square a negative or a possitive, both will give you the same result. Therefore, if you take the square root of the square, you end with either a possitive or negative number.

The concept of principal square root is not the same as a square root. It only considers the possitive solution. It isn't stated on the problem that we must consider that.

Rafael Acosta - 5 years, 7 months ago

@Rafael Acosta – Because it doesn't have to state it, you are required to know it. In Algebra 1, as soon as square root is discussed, you are told the Principal Square Root rule. Long before working on expressions with roots and variables, you are told that, for example, if you see

$\sqrt{25}$

the result is 5. NOT $\pm5$

Just 5.

In order to get -5, you have to write

$-\sqrt{25}$

Principal square root is expected. You're expected to know that. Your math teachers didn't drill that enough. They didn't drill that

$\sqrt{x^{2}}=|x|$

enough either. It's not your fault. But you do need to learn it now.

Pamela Barnes - 5 years, 7 months ago

@Rafael Acosta – $\sqrt{x^2}$ is defined to be $|x|$ in order to ensure that the square root function remains a function.

Abdur Rehman Zahid - 5 years, 7 months ago

I guess your sequence is wrong... When we put square for sign both sides, +-came out of the root.

Shreyansh Ajmera - 5 years, 7 months ago

@Shreyansh Ajmera – Unfortunately, your math teachers did not serve you well, if you are not aware that

$\sqrt{x^2}=|x|$

When a square root is taken, the resulting expression must be positive, due to the Principal Square Root rule. That's the purpose of the absolute value symbols - to guarantee a positive expression. (Actually, when any even root is taken, the result must be positive.)

So,

$\sqrt{x^2}=|x|$

$\sqrt{x^4}=|x^2|$

$\sqrt{x^6}=|x^3|$

$\sqrt{x^8}=|x^4|$

Now,

$|x|\neq x$ when $x<0$

$|x^{2}|$ = $x^{2}$ for all real x

$|x^{3}|\neq$ $x^{3}$ when $x<0$

$|x^{4}|$ = $x^{4}$ for all real x

And THAT is why only statements 2 and 4 are true.

Pamela Barnes - 5 years, 7 months ago

if x=1 X^2=1 sqrt of 1 is still 1 1 works for all of them please explain

matt kemp - 5 years, 7 months ago

Try using -1 and see if they all still work.

Pamela Barnes - 5 years, 7 months ago

isnt it just asking if any numbers work for this? not multiple numbers

matt kemp - 5 years, 7 months ago

@Matt Kemp – No, it's asking which are true for all real numbers. (That's what it means when it's phrased, "if x is a real number"). So it's not good enough to find one instance when it works. You have to know it works for any real number chosen. That's why I suggested you try -1, or even better -2, just because 1 is special sometimes. You'll see that if you use a negative number, only statements 2 and 4 work.

Pamela Barnes - 5 years, 7 months ago

@Pamela Barnes – ok thank you i thought it was asking if there are any real numbers that work.

matt kemp - 5 years, 7 months ago

Thomas James Bautista
Nov 8, 2015

Statement 1:

$x=-1$

$\sqrt{(-1)^2} \stackrel{?}{=} (-1)$

$\sqrt{1} \stackrel{?}{=} -1$

$1 \neq -1$

Therefore, statement 1 is false.

Statement 3:

$x=-1$

$\sqrt{(-1)^6} \stackrel{?}{=} (-1)^3$

$\sqrt{1} \stackrel{?}{=} -1$

$1 \neq -1$

Therefore, statement 3 is false.

By process of elimination, the only choice left is "Statement 2 and Statement 4"

Efren Medallo
Nov 1, 2015

Since it's only stated in the problem that $x$ is a real number, statements 1 and 3 will not hold true for negative values of $x$ , as the LHS will be positive, while the RHS will be negative.

Abhishek Dass
Nov 7, 2015

The square root function gives only positive root. Since even powers of x are positive, those statements are true

Please see Pamela Barnes' reply above.

Sabrina Q - 5 years, 7 months ago

This is true, and as such all four statements are correct.

David Taylor - 5 years, 7 months ago

No, I meant statements 2&4 are correct because odd powers have a possibility of being negative.

Abhishek Dass - 5 years, 7 months ago

Since the square root function only returns positive roots, I'm saying all four are true. Root(x^2) is not the same as the solutions for x^2. The solutions will give + and - values, whereas the root will only give the positive.

David Taylor - 5 years, 7 months ago

@David Taylor – I still don't understand why you say all four are true, then?

Glen Isip - 5 years, 7 months ago

@David Taylor – Square root of x^2 will give |x| since we don't know x is positive or negative. So we take it's modulus. Suppose x is -1. X^2 is 1. Now take it's square root. Since this function returns positive root, it gives 1. This whole thing is left hand side of statement 1. Right hand side is x which is in this case -1. Since RHS and LHS are not equal, statement is false. Hope you get it now.

Abhishek Dass - 5 years, 7 months ago

@Abhishek Dass – I realised this about an hour ago. I was only going one way, and seeing root(x^2) = x and not x = root(x^2). Unfortunately, writing on my phone, I couldn't redact my previous statements as I couldn't find my way back. First my maths, then my understanding of technology... What next?!

David Taylor - 5 years, 7 months ago

Chinmay Chhajed
Nov 9, 2015

For this question it should be known that sqrt(x) = |x|; not x

I believe you meant sqrt(x^2) = |x|.

Akeel Howell - 5 years ago

Dylan Ruff
Mar 7, 2016

While the mathematical explanation is shown I'll demomstrate with one value. X=-1 1) Sqrt (x^2) = x Sqrt (-1^2) = -1 Sqrt (1) =-1 1 doesn't not equal -1

2) sqrt (x^4) = x^2 Sqrt (-1^4) = -1^2 Sqrt (1) = 1 1=1

3) similar to 1

4) similar to 2.

Julie B.
Mar 6, 2016

The question would be better phrased as which of the statements above are always true. Statements 1 and 3 can be true in some cases but not all.

Aarabdh Tiwari
Jan 7, 2016

See here. 1. Even powers always produce positive number. 2. \sqrt(x^{2}) is always equal to positive value of x, rather than all values that satisfy x.(in short, modulo(x)) 3. Its NOT the same as x^{2} - y^{2}=0. In this case, there will be 2 values satisfying x, i.e., ±y. For those of you who didn't realise the difference between 2. and 3., kindly solve more problems based on this, to fully clarify the doubt.

Confusing square roots

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