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If x < 0, then root({x} •x) is [#permalink]
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If x < 0, then \(\sqrt{x*x}\)) is A. x B. 1 C. 1 D. x E. \(\sqrt{x}\)
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Last edited by Bunuel on 04 Dec 2012, 01:22, edited 2 times in total.
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Re: Square root and Modulus [#permalink]
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udaymathapati wrote: If x < 0, then \sqrt{x} •x) is A. x B. 1 C. 1 D. x E. \sqrt{x} 1. If x<0, then \(\sqrt{x*x}\) equals:A. \(x\) B. \(1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\) Remember: \(\sqrt{x^2}=x\). The point here is that square root function can not give negative result: wich means that \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function! That is why \(\sqrt{x^2}=x\) Back to the original question:\(\sqrt{x*x}=\sqrt{(x)*(x)}=\sqrt{x^2}=x=x\) Or just substitute the value let \(x=5<0\) > \(\sqrt{x*x}=\sqrt{(5)*5}=\sqrt{25}=5=(5)=x\). Answer: A. Hope it's clear.
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Re: Square root and Modulus [#permalink]
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02 Sep 2010, 13:07
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Re: Square root and Modulus [#permalink]
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02 Sep 2010, 13:14
Thanks, the way you posted the question clarified the solution. The original representation was quite confusing



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Re: Square root and Modulus [#permalink]
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06 Sep 2010, 15:36
Thanks bunuel, what I would do without your explanations!!!
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Re: Square root and Modulus [#permalink]
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21 Sep 2010, 03:14
Excellent post Bunuel +1



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Re: Square root and Modulus [#permalink]
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22 Sep 2010, 09:56
Thanks to Bunuel. Key thing here is Remember: Square root of a number on gmat is positive number I had seen similar problem earlier on gmatclub where I learnt this and got this one answer correct
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03 Nov 2010, 02:01
Thanks Bunuel for nice explanation.
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If x < 0, then root({x} •x) is [#permalink]
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shrive555 wrote: Bunel: i've already seen all the explanation just one more question.
as If \(x<0\), then \(\sqrt{x*x}\) Ans is x is Answer of the question depends on the condition x<0 or it depends on the sqrt (even root)
lets keep the condition same i.e x<0 and take odd root say cube root. i.e \(\sqrt[3]{x*x}\) . what would be the answer, would it be x then ?
Thanks About the odd roots: odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\). So, if given that \(x<0\) then \(x=x\) and \(x*x=(x)*(x)=positive*positive=x^2\), thus odd root from positive \(x^2\) will be positive. But \(\sqrt[3]{x*x}\) will equal neither to x nor to x: \(\sqrt[3]{x*x}=\sqrt[3]{x^2}=x^{\frac{2}{3}}\), for example if \(x=8<0\) then \(\sqrt[3]{x*x}=\sqrt[3]{x^2}=\sqrt[3]{64}=4\). If it were \(x<0\) and \(\sqrt[3]{x^2*x}=?\), then \(\sqrt[3]{x^2*x}=\sqrt[3]{x^2*(x)}=\sqrt[3]{x^3}=x<0\). Or substitute the value let \(x=5<0\) > \(\sqrt[3]{x^2*x}=\sqrt[3]{25*5}=\sqrt[3]{125}=5=x<0\). Now, back to the original question:If x<0, then \(\sqrt{x*x}\) equals:A. \(x\) B. \(1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\) As square root function cannot give negative result, then options 1 (B) and x (D) can not be the answers as they are negative. Also \(\sqrt{x}\) (E) can not be the answer as even root from negative number is undefined for GMAT. 1 (C) also can not be the answer as for different values of x the answer will be different, so it can not be some specific value. So we are left with A. Now, if it were x>0 instead of x<0 then the question would be flawed as in this case the expression under the even root would be negative (x*x=negative*positive=negative). Hope it's clear.
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Re: Square root and Modulus [#permalink]
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08 Dec 2010, 22:10
Excellent !!! THanks B
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Re: Square root and Modulus [#permalink]
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Friends, In GMAT, to avoid the confusion with the negative #s, i use my own and simple technique that is as below. PLEASE NOTE THAT: DEALING WITH THE INEQUALITIES/SQUAREROOTS/MODULUS THAT CONTAIN +VE #s IS EASIER THAN DEALING WITH THOSE CONTAIN VE#S. Given: x is a ve I take: y as a +ve #. now i write x = y FROM NOW ON I AM GONNA DEAL WITH THE +VE # (i.e.y) ONLY. I CAN NOT BE TRAPPED BY THE TESTMAKER. qtn: \(\sqrt{x*x}\) = ? ==> \(\sqrt{y*y}\) ==> \(\sqrt{y*y}\) as #=# ==> \(\sqrt{y^2}\) ==> \(y\) (but not y as Y is a positive #) ==> \(x\) Answer "A" Refer cananyoneanswerthisdsquestion94854.html for q qtn that can be solved using this technique. Regards, Murali. Kudos.



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Re: Square root and Modulus [#permalink]
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09 Dec 2010, 09:03
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Murali: THe concept is same \(\sqrt{x*x}\) given x<0=> \(\sqrt{x^2}\) = x x = x if x<0 x = x if x>0 since x<0 so x
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Re: Square root and Modulus [#permalink]
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23 Dec 2010, 18:04
great post Bunuel, thanks for the explanation.



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Re: Square root and Modulus [#permalink]
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30 Dec 2010, 14:20
marijose wrote: I still do not understend why the x is negative. Is it because of the condition at the begining or because of the absolute value that is in the question?
here is whay I see. SQ ( (x)) *x
you get... sq(x*x) therefore the answer is x, do you add the negative sign because the problem starts out saying that x is negative or is there a rule I am not seeing?
I was under the impression that the square root of any number cannot by negative or is this incorrect?
regards x is negative because it's given in the stem that it's negative (if x<0...). Next, knowing that x is negative doesn't mean that you should write x instead of x. As for the square root: yes, even roots can not give negative result, so when you say that \(\sqrt{x*x}\) equals to x, you are basically saying that square root of an expression equals to the negative number (as x is negative) which can not be true. Now, the answer is x or (negative)=positive, so square root of an expression equals to positive number which can be true. Also \(\sqrt{x^2}\) equals to x not to x. It's all explained above, so please read the solutions again and ask if anything needs further clarification. You can also check Absolute Value ( mathabsolutevaluemodulus86462.html) and Number Theory ( mathnumbertheory88376.html) chapters of Math Book for fundamentals.
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31 Dec 2010, 08:06
shrive555 wrote: marijose wrote: Ive read the explanations .. sorry to keep bugging but
lets say you plug in a a 5. by doing all the calculations you get that the SQ((5)*5) = sq(5*5) = 5 so because and only because the question states that x is negative we have to change the value to 5. I this the correct reasoning? you don't have to plug in 5. Just Plug in 5, keep in mind that x<0 so it will make 5 as 5 automatically. so when you get the final answer after taking square root of 25 = 5. due to given condition of x<0 . you have to pick 5 or (negative value)from the answer choices. The only trick is " x<0 "
i was stuck in the same point What do you mean by "it will make 5 as 5 automatically"??? And again: given that \(x<0\). If you want to solve this question by number plugging method you should plug negative values of \(x\) > if \(x=5<0\) then \(\sqrt{x*x}=\sqrt{(5)*5}=\sqrt{25}=5\) so the answer is 5 or \(x\), as \(x=(5)=5\).
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Re: Square root and Modulus [#permalink]
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31 Dec 2010, 10:33
Bunuel wrote: shrive555 wrote: marijose wrote: Ive read the explanations .. sorry to keep bugging but
lets say you plug in a a 5. by doing all the calculations you get that the SQ((5)*5) = sq(5*5) = 5 so because and only because the question states that x is negative we have to change the value to 5. I this the correct reasoning? you don't have to plug in 5. Just Plug in 5, keep in mind that x<0 so it will make 5 as 5 automatically. so when you get the final answer after taking square root of 25 = 5. due to given condition of x<0 . you have to pick 5 or (negative value)from the answer choices. The only trick is " x<0 "
i was stuck in the same point What do you mean by "it will make 5 as 5 automatically"??? And again: given that \(x<0\). If you want to solve this question by number plugging method you should plug negative values of \(x\) > if \(x=5<0\) then \(\sqrt{x*x}=\sqrt{(5)*5}=\sqrt{25}=5\) so the answer is 5 or \(x\), as \(x=(5)=5\). Since x=5 but x<0 so x=5probably marijose is confusing the negative sign  with negative x .... \(\sqrt{ (x)*x}\)
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31 Dec 2010, 11:31
Bunuel wrote: marijose wrote: I still do not understend why the x is negative. Is it because of the condition at the begining or because of the absolute value that is in the question?
here is whay I see. SQ ( (x)) *x
you get... sq(x*x) therefore the answer is x, do you add the negative sign because the problem starts out saying that x is negative or is there a rule I am not seeing?
I was under the impression that the square root of any number cannot by negative or is this incorrect?
regards x is negative because it's given in the stem that it's negative (if x<0...). Next, knowing that x is negative doesn't mean that you should write x instead of x.This is what i meant but messed up in mathematics lingo
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Re: Is the square root always positive? [#permalink]
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Merging similar topics. karenhipol wrote: I encountered this question in gmat prep:
If x is < 0, what is the Square root of x/x/
a. 0 b. 1 c. x d. x e. 1
The official answer is x but I answered x because I thought the answer should be positive.
Should I consider however, that since x <0 then I should multiply it by 1 to get a positive number so the answer is  x ( to get (x) = +x.
Please let me know If I reasoned this out correctly.
Thanks Two things: 1. Square root function can not give negative result: \(\sqrt{some \ expression}\geq{0}\) so square root of a nonnegative number is not always positive, it's never negative (always nonnegative). 2. As \(x<0\) then \(x=negative=positive\) so the answer is still positive. For the complete solution refer to the above posts.
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