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Fremontian
Joined: 14 Oct 2009
Last visit: 24 Mar 2016
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Posts: 21
Updated on: 15 Aug 2017, 21:53
Originally posted by Fremontian on 14 Dec 2009, 16:58.
Last edited by Bunuel on 15 Aug 2017, 21:53, edited 2 times in total.
Last edited by Bunuel on 15 Aug 2017, 21:53, edited 2 times in total.
Renamed the topic and edited the question.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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If \(x\neq{0}\), then \(\frac{\sqrt{x^2}}{x}=\)
A. -1
B. 0
C. 1
D. x
E. \(\frac{|x|}{x}\)
A. -1
B. 0
C. 1
D. x
E. \(\frac{|x|}{x}\)
If \(x\neq{0}\), then \(\frac{\sqrt{x^2}}{x}=\)
A. -1
B. 0
C. 1
D. x
E. \(\frac{|x|}{x}\)
General rule: \(\sqrt{x^2}=|x|\).
When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) cannot be negative but \(x\) can.
\(y\) cannot be negative as \(y=\sqrt{some \ expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).
When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.
That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.
Even roots have only a positive value on the GMAT.
Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.
In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that \(x\) is positive then the answer would be 1, if we knew that \(x\) is negative the answer would be -1. BUT we don't know the sign of x, hence we cannot simplify expression \(\frac{|x|}{x}\) further.
Hope it's clear.
_________________
A. -1
B. 0
C. 1
D. x
E. \(\frac{|x|}{x}\)
General rule: \(\sqrt{x^2}=|x|\).
When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=|x|\), which means that \(y\) cannot be negative but \(x\) can.
\(y\) cannot be negative as \(y=\sqrt{some \ expression}\), and even root from the expression (some value) is never negative (as for GMAT we are dealing only with real numbers).
When the GMAT provides the square root sign for an even root, such as a square root, then the only accepted answer is the positive root.
That is, \(\sqrt{16} = 4\), NOT +4 or -4. In contrast, the equation x^2 = 16 has TWO solutions, +4 and -4.
Even roots have only a positive value on the GMAT.
Odd roots will have the same sign as the base of the root. For example, CBRT(64) = 4, CBRT(-27) = -3.
In our original question we have: \(\frac{\sqrt{x^2}}{x}=\frac{|x|}{x}\), if we knew that \(x\) is positive then the answer would be 1, if we knew that \(x\) is negative the answer would be -1. BUT we don't know the sign of x, hence we cannot simplify expression \(\frac{|x|}{x}\) further.
Hope it's clear.
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Fremontian
For ANY value of \(x>0\) --> \(\frac{|x|}{x}=1\)
For ANY value of \(x<0\) --> \(\frac{|x|}{x}=-1\)
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General Discussion
Fremontian
Joined: 14 Oct 2009
Last visit: 24 Mar 2016
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Posts: 21
Thank you. Much clear now.
Also,
Since x is Not equal to zero, is there a value of x for which this equation \(\frac{|x|}{x}\) WILL NOT simplify to \(1\)?
Also,
Since x is Not equal to zero, is there a value of x for which this equation \(\frac{|x|}{x}\) WILL NOT simplify to \(1\)?
whiplash2411
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Last visit: 02 Mar 2015
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Schools: HBS '17 (M) Stanford '17 (A)
Schools: HBS '17 (M) Stanford '17 (A)
Posts: 1,760
You are asked to find the value of \(\frac {\sqrt{x^2}}{x}\)
Now, for the numerator term:
\(\sqrt{x^2} = x\) or \(-x\)
And the denominator is x.
So if you take the positive value of x as the numerator, the answer is 1, and if you take the negative value, the answer is -1. However, the root symbol specified is used only for positive answers.
Hence the only way to account for this is by using the mod sign. Hence the answer is E.
D is wrong, because either way you look at it \(\frac {\sqrt{x^2}}{x}\) can only be + or - 1
Hope this helps!
Now, for the numerator term:
\(\sqrt{x^2} = x\) or \(-x\)
And the denominator is x.
So if you take the positive value of x as the numerator, the answer is 1, and if you take the negative value, the answer is -1. However, the root symbol specified is used only for positive answers.
Hence the only way to account for this is by using the mod sign. Hence the answer is E.
D is wrong, because either way you look at it \(\frac {\sqrt{x^2}}{x}\) can only be + or - 1
Hope this helps!
Posts: 57
Lets try for say x=5;
sqroot(5^2)/5 = +5/5 or -5/5 = 1 or -1
so, "a", "b" and "c" is not right.
|x|/x = |5|/5 = 1
Also, for negative value of x, (say -5), |x|/x = |-5|/-5 = 5/-5 = -1
It looks ok, however;
|x|/x = 1 only for x>0;
|x|/x = -1 only for x<0;
which not the case with sqroot(x^2)/x (as, we just saw even for positive value of x, we can have it's value as -1).
So, "e" should be the right option.
Thanks
sqroot(5^2)/5 = +5/5 or -5/5 = 1 or -1
so, "a", "b" and "c" is not right.
|x|/x = |5|/5 = 1
Also, for negative value of x, (say -5), |x|/x = |-5|/-5 = 5/-5 = -1
It looks ok, however;
|x|/x = 1 only for x>0;
|x|/x = -1 only for x<0;
which not the case with sqroot(x^2)/x (as, we just saw even for positive value of x, we can have it's value as -1).
So, "e" should be the right option.
Thanks
whiplash2411
Joined: 09 Jun 2010
Last visit: 02 Mar 2015
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Status:Three Down.
Concentration: General Management, Nonprofit
Schools: HBS '17 (M) Stanford '17 (A)
Schools: HBS '17 (M) Stanford '17 (A)
Posts: 1,760
You are basically asked to find out what \(\frac{\sqrt{x^2}}{x}\) is.
If it had been given as \(\frac{\sqrt{x}}{x}\) then we can say that the numerator is simply x, and neglect the -x value since the square root sign considers only the positive radical.
But judging by the OA, I think the question was given in terms of what I had written in the first statement. In that case, the numerator is either a +x or a -x, depending on the original value of x. But since we don't know whether the original value was a + or a - number, we use the mod sign to indicate that we are taking the absolute value of the number, which is always positive. So your final answer will be \(\frac{|x|}{x}\)
As an example, let's consider one positive and one negative case.
x = 1
\(x^2\)= 1
\(\sqrt{x^2}\) = x = 1
So here, \(\frac{\sqrt{x^2}}{x}\) = 1
This poses no confusion since the original x value was a positive number by itself.
x = -1
\(x^2\) = 1
\(\sqrt{x^2}\) = (-x) = 1 [Note: The radical sign only indicates that the final result has to be a positive number. This doesn't necessarily mean that the answer is always 'x']
So here, we have \(\frac{\sqrt{x^2}}{x}\) = \(\frac{-x}{x}\) = -1
So, to combine both these results into one answer that fits both, we use \(\frac{|x|}{x}\)
Hope this helps.
If it had been given as \(\frac{\sqrt{x}}{x}\) then we can say that the numerator is simply x, and neglect the -x value since the square root sign considers only the positive radical.
But judging by the OA, I think the question was given in terms of what I had written in the first statement. In that case, the numerator is either a +x or a -x, depending on the original value of x. But since we don't know whether the original value was a + or a - number, we use the mod sign to indicate that we are taking the absolute value of the number, which is always positive. So your final answer will be \(\frac{|x|}{x}\)
As an example, let's consider one positive and one negative case.
x = 1
\(x^2\)= 1
\(\sqrt{x^2}\) = x = 1
So here, \(\frac{\sqrt{x^2}}{x}\) = 1
This poses no confusion since the original x value was a positive number by itself.
x = -1
\(x^2\) = 1
\(\sqrt{x^2}\) = (-x) = 1 [Note: The radical sign only indicates that the final result has to be a positive number. This doesn't necessarily mean that the answer is always 'x']
So here, we have \(\frac{\sqrt{x^2}}{x}\) = \(\frac{-x}{x}\) = -1
So, to combine both these results into one answer that fits both, we use \(\frac{|x|}{x}\)
Hope this helps.
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I think that A and C are out because \(\sqrt{x^2}/x\) could be +1 and -1.
Remember that the square root of a number can have a positive and negative value.
E makes sense, but let's wait a better explanation.
Remember that the square root of a number can have a positive and negative value.
E makes sense, but let's wait a better explanation.
Posts: 109
satishreddy
Square root of (x^2) ,can be +x or -x and this can be written as |x|.Because Modulus of anything should be always positive as we are just takking the magnitude.If x is +ve ,|x| will be x as its +ve.If x is -ve ,|x| will be -x
as -ve of -ve number will be +ve.
Hence teh answer will be |x|/x
Consider KUDOS if it is helpful to u in some way.
Other questions about the same concept:
if-x-81600.html
square-root-and-modulus-100303.html
Hope it helps.
_________________
if-x-81600.html
square-root-and-modulus-100303.html
Hope it helps.
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if x>0 then sqrt(x^2) = x
if x<0 then sqrt(x^2) = -x
so we generalize the given expression sqrt(x^2)/x = |x|/x
Answer is E.
if x<0 then sqrt(x^2) = -x
so we generalize the given expression sqrt(x^2)/x = |x|/x
Answer is E.
Posts: 3,651
naeln
Hi naeln,
\(\sqrt{x^2} = |x|\) i.e. square root function will not give a negative value. Let me explain this to with you an example.
Assume \(x = 2\), so \(x^2 = 4\) and hence \(\sqrt{x^2} = \sqrt{4} = 2\) \(= x\) when \(x => 0\)
Similarly if \(x = -2, x^2 = 4\) and hence \(\sqrt{x^2} = \sqrt{4} = 2\) \(= - x\) when \(x < 0\)
So the square root function has the same behavior as the modulus function. Hence we need to represent \(\sqrt{x^2} = |x|\).
Hope it's clear
Regards
Harsh
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Bunuel
Hi Bunuel,
One question here. If x is positive then the answer would be 1. But if x is negitive means x<0. so |-x| = |x| or -(-x) = x which would mean the answer is 1.
asciijai
If x = -1, then \(\frac{|x|}{x}= \frac{|-1|}{-1}=\frac{1}{-1}=-1\)
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Another conceptual question:
If we have in a question square root of any variable (e.g. \(\sqrt{x^2}\)) then the answer would modulus x, provided no information is provided in the question stem.
If we have square root of any number (e.g. \(\sqrt{25}\)) then we take only the positive root, viz. 5 (even without any further information), because that's what the GMAT does. Nevertheless, normally it could be both 5 and negative 5.
Please confirm my understanding, because that's how I read this thread and the GMATClub Math Guide.
Thanks for the clarification!
If we have in a question square root of any variable (e.g. \(\sqrt{x^2}\)) then the answer would modulus x, provided no information is provided in the question stem.
If we have square root of any number (e.g. \(\sqrt{25}\)) then we take only the positive root, viz. 5 (even without any further information), because that's what the GMAT does. Nevertheless, normally it could be both 5 and negative 5.
Please confirm my understanding, because that's how I read this thread and the GMATClub Math Guide.
Thanks for the clarification!
bgpower
Everything is correct, except the red text.
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Bunuel
TITCR. Can't explain any better myself!
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asciijai
Hi asciijai,
|x| = +x when x > 0 i.e +ve
|x| = -x when x < 0 i.e -ve
|x|/x
If x = 5 then |5|/5 = 1
If x = -5 then |-5|/-5 = 5/-5 = -1
When you open the mod then it makes the value positive. so what is inside mod if that is positive then fine but if that is negative then you multiply that by negative to make the outcome +ve.
if x>0
|x| = +x so if x=5 then
|5| = +5
if x<0
|x| = -x so if x=-5 then
|-5| = -(-5) = 5
Hope this is clear.
Fremontian
(A) is the answer if x = -ve
(B) cannot be the answer since its given \(x\neq{0}\)
(C) is the answer if x = +ve
(D) I don't think this can ever be the answer
(E) ALWAYS TRUE
Bunuel
Experts -
To clarify, for the general rule mentioned above: x2−−√=|x|x2=|x|
X can be - or +, correct? When I saw just the absolute value bars, I was not sure if you should only be taking the positive solution or if the absolute value bars mean you need to then solve for the absolute value by having one positive and one negative case for x.
Many thanks!
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