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# If x#0, then root(x^2)/x=

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14 Dec 2009, 17:58
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If $$x\neq{0}$$, then $$\frac{\sqrt{x^2}}{x}=$$

A. -1
B. 0
C. 1
D. x
E. $$\frac{|x|}{x}$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Dec 2012, 03:14, edited 1 time in total.
Renamed the topic and edited the question.
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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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Re: If x does not equal 0, then root x^2/x = [#permalink]

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28 May 2015, 23:59
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Expert's post
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: What is sqrt of x^2? [#permalink]

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15 Dec 2009, 04:28
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Expert's post
Fremontian wrote:
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 $$1$$?

For ANY value of $$x>0$$ --> $$\frac{|x|}{x}=1$$
For ANY value of $$x<0$$ --> $$\frac{|x|}{x}=-1$$
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19 Oct 2010, 14:16
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Expert's post
Other questions about the same concept:
if-x-81600.html
square-root-and-modulus-100303.html

Hope it helps.
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Re: If x#0, then root(x^2)/x= [#permalink]

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15 Jun 2015, 02:54
1
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Expert's post
bgpower wrote:
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.

Thanks for the clarification!

Everything is correct, except the red text.
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Re: What is sqrt of x^2? [#permalink]

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14 Dec 2009, 18:32
The problem is we can't simply assume that because \sqrt{x} can't be negative it isn't. Instead, we have to make sure x isn't negative through use of absolute value.

Reading further in the OG, "Every positive number n has two square roots, one positive and the other negative...The two square roots of 9 are 3 and -3."

In $$x^2$$, x could be positive or negative for the equation to make sense, but to adhere to the rule you correctly stated, we use absolute value to ensure x is positive.
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Re: What is sqrt of x^2? [#permalink]

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14 Dec 2009, 19:41
Bunuel wrote:
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 can not simplify expression $$\frac{|x|}{x}$$ further.

Hope it's clear.

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: What is sqrt of x^2? [#permalink]

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14 Dec 2009, 20:50
Fremontian wrote:
Bunuel wrote:
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 can not simplify expression $$\frac{|x|}{x}$$ further.

Hope it's clear.

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 $$1$$?

if x=-1 then the exp will -1
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Re: What is sqrt of x^2? [#permalink]

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14 Dec 2009, 22:06
xcusemeplz2009 wrote:
Fremontian wrote:
Bunuel wrote:
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 can not simplify expression $$\frac{|x|}{x}$$ further.

Hope it's clear.

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 $$1$$?

if x=-1 then the exp will -1

duh! that was a stupid question from me. Thanks for clarifying!
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Re: So easy but couldnt do that explain me plz [#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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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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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: Square Root - Absolute Value [#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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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

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Re: What is sqrt of x^2? [#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

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03 Dec 2012, 20:40
I'll just remember that $$\sqrt{x^2}=|x|$$

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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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If x does not equal 0, then root x^2/x = [#permalink]

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28 May 2015, 23:29
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?
If x does not equal 0, then root x^2/x =   [#permalink] 28 May 2015, 23:29

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