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Square Root = Always Positive?

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New post 23 May 2011, 14:34
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My Princeton Review teacher said something in class that made me scratch my head. We were going over a DS problem, and the prompt was something along the lines of "What is the value of x?"

One of the statements read: \sqrt{\(x=16\)}

I said insufficient, because x could equal 4 or -4. My instructor said this incorrect, and that in instances like this, the GMAT will always assume the square root to be the positive root. Is this true on the actual GMAT? It doesn't seem to make sense to me. The last thing I want to do is get an easy answer wrong because of something like this.
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New post 23 May 2011, 14:52
Yes, it will always be the positive number. The point is that, by convention, square root refers to the principal square root.

\(x^2 = 9\)
\(\sqrt{x^2} = 3\)

This doesn't suggest that a solution to the equation isn't either
\(-3\) or \(3\)
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New post 23 May 2011, 17:15
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ebonn101 wrote:
My Princeton Review teacher said something in class that made me scratch my head. We were going over a DS problem, and the prompt was something along the lines of "What is the value of x?"

One of the statements read: \sqrt{\(x=16\)}

I said insufficient, because x could equal 4 or -4. My instructor said this incorrect, and that in instances like this, the GMAT will always assume the square root to be the positive root. Is this true on the actual GMAT? It doesn't seem to make sense to me. The last thing I want to do is get an easy answer wrong because of something like this.



It is true.

\(x^2 = 16\)
has two solutions: x = 4 or -4

\(x = \sqrt{16}\)
has only one solution: x = 4
'the square root' is used to refer to only the positive square root.
\(\sqrt{x^2} = |x|\)
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New post 23 May 2011, 19:03
You know what, I had the same confusion too. Thanks for clearing that up guys! Appreciate it
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New post 12 Oct 2011, 12:38
Hi Karishma,

Got a few questions for you:

What is the difference between the two questions:
Quote:
\(x^2 = 16\)

and
Quote:
\(x = \sqrt{16}\)


You say that square root only refers to the positive, yet for the first one you say it has 2 solutions. Little confused.

The way I see it, 4 is the solution to both questions in terms of the GMAT.

Secondly,

How is the following different from the above:
Quote:
\((x+1)^2 <= 36\)


I know that we are suppose to take the absolute value of the expression on the left: |x+1| <= 36

Why are we considering the negative here?
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Re: Square Root = Always Positive?  [#permalink]

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New post 13 Oct 2011, 10:04
386390 wrote:
Hi Karishma,

Got a few questions for you:

What is the difference between the two questions:
Quote:
\(x^2 = 16\)

and
Quote:
\(x = \sqrt{16}\)


You say that square root only refers to the positive, yet for the first one you say it has 2 solutions. Little confused.

The way I see it, 4 is the solution to both questions in terms of the GMAT.

\(x^2 = 16\) has 2 solutions. x = 4 and -4. Both of them satisfy the equation.
When you take the square root of this equation, you get \(\sqrt{x^2} = \sqrt{16}\)
Remember, \(\sqrt{x^2}\) is not x. It is |x| i.e. just the positive value.
x, on the other hand, can be positive or negative. So you get two values for x.

You get |x| = 4 (Only positive square root)
Since |x| = 4, x is either 4 or -4 (same as before)

This is different from \(x = \sqrt{16}\)
Here x = 4 only

Secondly,

How is the following different from the above:
Quote:
\((x+1)^2 <= 36\)


I know that we are suppose to take the absolute value of the expression on the left: |x+1| <= 36

In this inequality, both the LHS and RHS are positive. So we can take the square root.
\(\sqrt{(x+1)^2} <= \sqrt{36}\)
|x+1| <= 6
Again, x+1


Why are we considering the negative here?

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Re: Square Root = Always Positive?  [#permalink]

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New post 17 Oct 2011, 07:43
Hello guys,
Here is a question I asked on Y! Answers, I got many kinds of answers and I'm not sure anymore.
goo.gl/jPTJk
(Copy paste the link to your browser address bar)

I just want to know how x^2 = 49 gives us x= +-7

This is the way I did it and asked for opinions on Y! Answers, got too many different types of comments (you can check the link
goo.gl/jPTJk )
I did a google search and landed in this forum. Hope you guys help me :) or should i open a new thread on this? let me know :)
x^2 = 49
√ (x^2) = √ 49
|x| = |√ (7*7)| or |x| = |√ (-7*-7)|
|x| = |7| or |x| = |-7|
|x| = 7
x= +- 7
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Re: Square Root = Always Positive?  [#permalink]

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New post 17 Oct 2011, 22:20
greatiam wrote:
x^2 = 49
√ (x^2) = √ 49

Since √ 49 = 7 (because when there is a square root, you consider only the positive value), you get
|x| = 7 (why is there a mod? because you consider the square root to be positive)
Now what values can x take?
x = +-7

Think of it in another way:

x = √ 49 = 7 (only the positive value)
x^2 = 49 gives two values x = 7 or -7

Even though the two equations are equivalent, their intention is different.

x = 7
Squaring both sides, x^2 = 49 (which holds)

Now if you take the square root again, you don't get just x = 7. You get x = -7 too.
So you have to be careful when you take the square root.
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New post 18 Oct 2011, 08:43
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So basically,

\(\sqrt{x^2}\) = |x| ***is the golden rule. Anytime you take a root of a square, the other side will be in absolute form.

Hence the earlier example:
\(x^2 = 49\)
\(\sqrt{x^2} = \sqrt{49}\)
x = |7|
and since the 7 is in an absolute value form, behind the scenes, it can be a + or a -.
But even though it can be a - or a +, since theres a mod and in terms of the gmat, x = 7.
Correct?
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New post 18 Oct 2011, 09:19
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I think IanStewart also clarifies it by saying:

Quote:
There is a difference between talking about 'a square root of 16', and talking about \(\sqrt{16}\) . The square root ('radical') symbol means the non-negative square root. So while it's certainly true that 16 has two square roots, 4 and -4 [highlight](as explained by Karishma above)[/highlight], if you ever see \(\sqrt{16}\) , this is always equal to 4 and only 4, because of the definition of the square root symbol.


found at

Quote:

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New post 02 Jul 2012, 19:37
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I stumbled across this thread and just wanted to post some additional info for students who are still scratching their heads wondering why sqrt(x^2)=|x| It's not just some random convention, in fact there is a very simple and logical explanation for why it's this way! So instead of simply memorizing that "formula" sqrt(x^2)=|x|, try to understand the reasoning behind it, then you will remember it always!

Let's consider our square root function f(x)=sqrt(x) and for a specific example we'll consider f(16)=sqrt(16)


So what are we looking for? The key word in the above sentence is "function". Let's quickly review the details of functions. As we all know, if you give an "input" to a function, it can give you only one output. So whatever value we plug in for x, the value that goes under the square root sign, we are looking for its output. So what is the output for this function? Ponder this question for a moment and you will see where the confusion typically arises. The typical student sees f(x)=sqrt(16) and they think "This is telling me to find the numbers so that when I square them it will give me 16! Oh, ok, both 4 and -4 do that!" While it's true that both 4 and -4 when squared give us 16 (these are called the roots of 16), the issue is that's not what the function f(x)=sqrt(x) is asking for. Remember: for a function, one input means one single output. Because of this, the output for the square root function is defined as the principle root, or the "positive root". If we got all the roots (4, -4) for that one input (16) it would no longer be considered a function so we have to "pick" one to define the square root function, and the simplest most obvious pick is the principle root, or positive root.


So I hope that makes sense and hopefully someone might read that far and now understand why it is the way it is. Remember, there's always a logical explanation when it comes to mathematics! (that's what makes it such a lovely subject). If you ever find yourself scratching your head in confusion, just dig a little deeper!
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New post 04 Jul 2012, 06:27
Yes, unless we are talking about square root of zero :)
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New post 03 Nov 2014, 11:27
These answers really helped clear up my confusion as to why a square root is always positive. However, could someone go a little further into why ALL even roots must be a positive number?

Thanks!
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New post 04 Nov 2014, 04:16
rebeccaeliscu wrote:
These answers really helped clear up my confusion as to why a square root is always positive. However, could someone go a little further into why ALL even roots must be a positive number?

Thanks!


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
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Hope it helps.
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Re: Square Root = Always Positive?  [#permalink]

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New post 04 Oct 2017, 18:27
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!
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Re: Square Root = Always Positive?  [#permalink]

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New post 04 Oct 2017, 19:53
MartinSC wrote:
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).
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Re: Square Root = Always Positive?  [#permalink]

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New post 05 Oct 2017, 05:55
Bunuel wrote:
MartinSC wrote:
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).



Thanks, but in this case the correct answer is considering the negative root because (C) -24 is the correct answer. If it considered the positive root, the correct answer would have been -16, which is not even among the choices.

Am I missing something here? This is an OG question.

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Re: Square Root = Always Positive?  [#permalink]

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New post 05 Oct 2017, 06:35
MartinSC wrote:
Bunuel wrote:
MartinSC wrote:
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).



Thanks, but in this case the correct answer is considering the negative root because (C) -24 is the correct answer. If it considered the positive root, the correct answer would have been -16, which is not even among the choices.

Am I missing something here? This is an OG question.

Rgds,



Here is a solution for that question:

\((x - 1)^2 = 400\) --> \(x-1=20\) or \(x-1=-20\) --> \(x-5=20-4=16\) or \(x-5=-20-4=-24\).

Answer: C.

Discussed here: https://gmatclub.com/forum/if-x-1-2-400 ... 44460.html
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Re: Square Root = Always Positive?  [#permalink]

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New post 08 Oct 2017, 12:42
MartinSC wrote:
Bunuel wrote:
MartinSC wrote:
I was counting on that, but what about this one:

If (x-1)^2=400, which of the following could be the value of ?
(A) 15
(B) 14
(C) –24
(D) –25
(E) –26

I thought "16" applying the "always positive" mantra, but the explanation goes as follows:

"Work the problem by taking the square root of both sides and solving for x.

Thus,

(x-1)^2=400
x-1= +/-20 (wait, I was not expecting to see the negative root!!)

So x= -19 or x=21 and
x-5=-24 or x-5=16" (which is not even an option).

Seeing that 16 was not among the choices, I could have picked (C)-24, which would have been correct, but it goes against the theory of the positive root...so I was left scratching my head.

Anyone has an idea?

Thks!


When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).



Thanks, but in this case the correct answer is considering the negative root because (C) -24 is the correct answer. If it considered the positive root, the correct answer would have been -16, which is not even among the choices.

Am I missing something here? This is an OG question.

Rgds,


I'm glad you raised this question. It turns out that there are two different scenarios:

\(\sqrt{9}=x\)
\(9 = x^2\)

Those are two different things on the GMAT. This post goes into the specifics: https://www.manhattanprep.com/gmat/blog ... -the-gmat/
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Re: Square Root = Always Positive?  [#permalink]

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New post 11 Nov 2018, 19:23
ebonn101 wrote:
My Princeton Review teacher said something in class that made me scratch my head. We were going over a DS problem, and the prompt was something along the lines of "What is the value of x?"

One of the statements read: \sqrt{\(x=16\)}

I said insufficient, because x could equal 4 or -4. My instructor said this incorrect, and that in instances like this, the GMAT will always assume the square root to be the positive root. Is this true on the actual GMAT? It doesn't seem to make sense to me. The last thing I want to do is get an easy answer wrong because of something like this.


Recall that when the square root symbol √ is used, this is specific notation that ONLY the principal square root (i.e., the positive square root) is correct. Thus, we see that √16 = 4. You might have confused this with the solution to the equation x^2 = 16. The solution to the equation is x = 4 OR x = -4. In fact, the square root of a number y can be defined as the positive root of the equation x^2 = y.
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Re: Square Root = Always Positive? &nbs [#permalink] 11 Nov 2018, 19:23
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