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If x ≠ 0, then root(x^2)/x= [#permalink]
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Updated on: 15 Aug 2017, 22:53
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If \(x\neq{0}\), then \(\frac{\sqrt{x^2}}{x}=\) A. 1 B. 0 C. 1 D. x E. \(\frac{x}{x}\)
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Originally posted by Fremontian on 14 Dec 2009, 17:58.
Last edited by Bunuel on 15 Aug 2017, 22:53, edited 2 times in total.
Renamed the topic and edited the question.



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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14 Dec 2009, 18:47
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If \(x\neq{0}\), then \(\frac{\sqrt{x^2}}{x}=\)A. 1 B. 0 C. 1 D. x E. \(\frac{x}{x}\) General rule: \(\sqrt{x^2}=x\). When we see the equation of a type: \(y=\sqrt{x^2}\) then \(y=x\), which means that \(y\) can not be negative but \(x\) can. \(y\)can not 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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If x ≠ 0, then root(x^2)/x= [#permalink]
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14 Dec 2009, 22:06
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 \(1\)?



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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18 Jun 2010, 06:39
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!



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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05 Jul 2010, 18:19
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



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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06 Jul 2010, 06:23
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.



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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25 Aug 2010, 13:15
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.
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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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19 Oct 2010, 02:16
satishreddy wrote: need help 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.



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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19 Oct 2010, 14:16



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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19 Apr 2011, 19:17
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.



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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28 May 2015, 23:59
naeln wrote: If x does not equal 0, then\(\sqrt{x^2}\)/x = a. 1 b. 0 c. 1 d. x e. lxl/x
Could anyone explain why C is wrong (why couldn't we take the square root of \(\sqrt{x^2}\) so it would equal x) and how come E is right?! Also, what is the level of difficulty for a question like this? 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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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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04 Jun 2015, 01:24
Bunuel wrote: 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 \(y\) can not be negative but \(x\) can.
\(y\)can not 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. 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.



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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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04 Jun 2015, 02:50
asciijai wrote: Bunuel wrote: 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 \(y\) can not be negative but \(x\) can.
\(y\)can not 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. 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. If x = 1, then \(\frac{x}{x}= \frac{1}{1}=\frac{1}{1}=1\)
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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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15 Jun 2015, 02:52
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!
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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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25 Jun 2016, 07:53
Bunuel wrote: 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 \(y\) can not be negative but \(x\) can.
\(y\)can not 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. TITCR. Can't explain any better myself!
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Re: If x ≠ 0, then root(x^2)/x= [#permalink]
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05 Apr 2017, 11:51
asciijai wrote: 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. 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.
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