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If x<0, then root(x*x) is:
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29 Jul 2009, 16:42
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If x<0, then \(\sqrt{xx}\) is: A. x B. 1 C. 1 D. x E. \(\sqrt{x}\)
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Re: If x<0
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16 Sep 2010, 19:14
saxenashobhit wrote: hailtothethief23 wrote: It is actually A.
suppose x is 2 then you have sqrt(2*2) = sqrt(4) = 2 = x
note that x < 0 as otherwise the function does not exist. I had same question today \(\sqrt{4}\) = + or  2. So answer should be + or  x. I don't get how can OA be x SOME NOTES:1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. 2. Any nonnegative real number has a unique nonnegative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root. When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only nonnegative value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\). 3. \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). 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: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). BACK TO THE ORIGINAL QUESTION:1. If \(x<0\), then \(\sqrt{x*x}\) equals:A. \(x\) B. \(1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\) \(\sqrt{x*x}=\sqrt{(x)*(x)}=\sqrt{x^2}=x=x\). Note that as \(x<0\) then \(x=negative=positive\), so \(\sqrt{x*x}=x=positive\) as it should be. Or just substitute the some negative \(x\), 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: Question from GMAT Prep
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08 Jul 2010, 06:02
The easiest way for you to solve this problem would be to plug in a number and see what happens.
Let's say \(x = 1\)
\(\sqrt{xx}=\sqrt{(1)1}=\sqrt{(1)(1)}=1 = (1)\)
So, your answer is A, x.




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Re: If x<0
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02 Aug 2009, 18:14
It is actually A.
suppose x is 2 then you have sqrt(2*2) = sqrt(4) = 2 = x
note that x < 0 as otherwise the function does not exist.



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Re: Question from GMAT Prep
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08 Jul 2010, 07:29
No problem! In the future, when you see a problem like this, I would strongly suggest just plugging in values to see where that takes you. Especially if you don't get a hunch immediately upon looking at the problem, plug in values to get a feel for the problem. And then, if it feels conflicting or not entirely sufficient to answer, try a different number. Hope this helps!



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Re: Question from GMAT Prep
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16 Jul 2010, 15:58
Nice work whiplash! I agree 100% that plugging in numbers is often the most efficient (and least susceptible to careless errors) method on this kind of problem. One addt'l tipit can be a timesaver to avoid 0, 1 (& 1), and numbers in the answer choices on a problem solving question (as opposed to on data sufficiency question, when you actually WANT to find the exceptions to the rule) because those are special numbers with special properties (the result of squaring 0 and 1 is the same as the number you put in, which is unusual) . Notice that the result of plugging in 1 is 1, which is choice A but it is *also* choice C. If you start with a small prime number like 2 or 3 (or in this case 2's negative relative, 2) you can sometimes save yourself a second pass.
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Re: If x<0
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16 Sep 2010, 18:55
hailtothethief23 wrote: It is actually A.
suppose x is 2 then you have sqrt(2*2) = sqrt(4) = 2 = x
note that x < 0 as otherwise the function does not exist. I had same question today \(\sqrt{4}\) = + or  2. So answer should be + or  x. I don't get how can OA be x
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Re: If x<0
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16 Sep 2010, 20:14
I got that wrong, thanks "Bunuel"
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Re: If x<0
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16 Sep 2010, 21:15
Thanks. Only catch which I wasn't aware was that on square roots I should consider positive value alone. Although this is wrong based on general Math
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Re: If x<0
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Updated on: 06 Dec 2012, 05:57
Thank you for this post.
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Last edited by mbaiseasy on 06 Dec 2012, 05:57, edited 1 time in total.



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Re: If x<0
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04 Dec 2012, 00:46
nss123 wrote: I'm having trouble understanding how they get to the answer, though it should be very simple:
If x<0, then \(\sqrt{xx}\) is: A) x B) 1 C) 1 D) x E) \(\sqrt{x}\) I got D, b/c I'm thinking, if x is negative, then taking the opposite of that will be positive, mulitiplied by the absolute value of x, which is also positive. That gives you x^2, which you take the square root of, to get x.
Any help would be great. Square root of any number CANNOT be negative... The answer is quite apparent and should not take more than 10 seconds to figure out.. B & D can be immediately eliminated.. There is no evidence to show C.(C takes a constant value for the root of a variable) and E is an imaginary number... A is clearly the right answer..
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Re: If x<0, then root(x*x) is:
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06 Dec 2012, 05:56
Let x=5 \(\sqrt{(5)5}=\sqrt{25}=5\) 5 = (5) = x Answer: x or A
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Re: If x<0
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29 Aug 2013, 04:47
Bunuel wrote: saxenashobhit wrote: hailtothethief23 wrote: It is actually A.
suppose x is 2 then you have sqrt(2*2) = sqrt(4) = 2 = x
note that x < 0 as otherwise the function does not exist. I had same question today \(\sqrt{4}\) = + or  2. So answer should be + or  x. I don't get how can OA be x SOME NOTES:1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. 2. Any nonnegative real number has a unique nonnegative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root. When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only nonnegative value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\). 3. \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). 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: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). BACK TO THE ORIGINAL QUESTION:1. If \(x<0\), then \(\sqrt{x*x}\) equals:A. \(x\) B. \(1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\) \(\sqrt{x*x}=\sqrt{(x)*(x)}=\sqrt{x^2}=x=x\). Note that as \(x<0\) then \(x=negative=positive\), so \(\sqrt{x*x}=x=positive\) as it should be. Or just substitute the some negative \(x\), let \(x=5<0\) > \(\sqrt{x*x}=\sqrt{(5)*5}=\sqrt{25}=5=(5)=x\). Answer: A. Hope it's clear. When x is greater than equal to zero , x is positive when x is less than zero, x is negative. can we also say as below ? when x is greater than zero , x is positive, and when x is less than equal to zero, x is negative . The main point is can we switch equal condition on either side?



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Re: If x<0
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29 Aug 2013, 04:55
ygdrasil24 wrote: Bunuel wrote: saxenashobhit wrote: I had same question today
\(\sqrt{4}\) = + or  2. So answer should be + or  x. I don't get how can OA be x SOME NOTES:1. GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. 2. Any nonnegative real number has a unique nonnegative square root called the principal square root and unless otherwise specified, the square root is generally taken to mean the principal square root. When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or 5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and 5. Even roots have only nonnegative value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{64} =4\). 3. \(\sqrt{x^2}=x\). The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\). 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: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). BACK TO THE ORIGINAL QUESTION:1. If \(x<0\), then \(\sqrt{x*x}\) equals:A. \(x\) B. \(1\) C. \(1\) D. \(x\) E. \(\sqrt{x}\) \(\sqrt{x*x}=\sqrt{(x)*(x)}=\sqrt{x^2}=x=x\). Note that as \(x<0\) then \(x=negative=positive\), so \(\sqrt{x*x}=x=positive\) as it should be. Or just substitute the some negative \(x\), let \(x=5<0\) > \(\sqrt{x*x}=\sqrt{(5)*5}=\sqrt{25}=5=(5)=x\). Answer: A. Hope it's clear. When x is greater than equal to zero , x is positive when x is less than zero, x is negative. can we also say as below ? when x is greater than zero , x is positive, and when x is less than equal to zero, x is negative . The main point is can we switch equal condition on either side? Are you referring to this: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). If yes, then you can put = sign in either case.
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Re: If x<0
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29 Aug 2013, 09:06
Are you referring to this: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\).
If yes, then you can put = sign in either case.[/quote]
Yes I was asking for mod definition only, so how to decide which way to go , I mean
Defn1 : X = +x when x> 0, and x=x when x is less than eq to zero Defn2: X = +x when xis greater than eq 0, and x=x when x is less than zero
Which defn to go for and when, totally confused in this. Please adcvise.



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Re: If x<0
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29 Aug 2013, 09:51
ygdrasil24 wrote: Are you referring to this: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\).
If yes, then you can put = sign in either case. Yes I was asking for mod definition only, so how to decide which way to go , I mean Defn1 : X = +x when x> 0, and x=x when x is less than eq to zero Defn2: X = +x when xis greater than eq 0, and x=x when x is less than zero Which defn to go for and when, totally confused in this. Please adcvise.[/quote] It does NOT matter in which part you put = sign.
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Re: If x<0, then root(x*x) is:
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16 Apr 2016, 08:00
nss123 wrote: If x<0, then \(\sqrt{xx}\) is:
A. x B. 1 C. 1 D. x E. \(\sqrt{x}\) if x <0 then x = x so this way we get \(\sqrt{(x)*(x)}\) i.e. \(\sqrt{x^2}\) i.e. x. As already said, x = x, Option A is correct.
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Re: If x<0, then root(x*x) is:
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16 Apr 2016, 10:59
HKD1710 wrote: nss123 wrote: If x<0, then \(\sqrt{xx}\) is:
A. x B. 1 C. 1 D. x E. \(\sqrt{x}\) if x <0 then x = x so this way we get \(\sqrt{(x)*(x)}\) i.e. \(\sqrt{x^2}\) i.e. x. As already said, x = x, Option A is correct. This is incorrect. If x < 0, then we get \(\sqrt{ (x)*(x)}\) i.e. \(\sqrt{x^2}\)



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Re: If x<0, then root(x*x) is:
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16 Apr 2016, 11:21
atturhari wrote: HKD1710 wrote: nss123 wrote: If x<0, then \(\sqrt{xx}\) is:
A. x B. 1 C. 1 D. x E. \(\sqrt{x}\) if x <0 then x = x so this way we get \(\sqrt{(x)*(x)}\) i.e. \(\sqrt{x^2}\) i.e. x. As already said, x = x, Option A is correct. This is incorrect. If x < 0, then we get \(\sqrt{ (x)*(x)}\) i.e. \(\sqrt{x^2}\) Hi atturhari, \(\sqrt{x^2}\) is NOT CORRECT. lets say x=2. put this value in \(\sqrt{ (x)*(x)}\) and you get \(\sqrt{ ((2))*((2))}\) i.e. \(\sqrt{ (2)*(2)}\) i.e. \(\sqrt{ 4}\) , which is not defined. \(\sqrt{(x)*(x)}\) is correct. because x is negative and in this case first of all, you just place x in terms of x and i.e x. Multuply x with itself will give x^2. now square root of x^2 is nothing but x which is x in our case.
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Re: If x<0, then root(x*x) is:
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16 Apr 2016, 11:41
atturhari wrote: HKD1710 wrote: nss123 wrote: If x<0, then \(\sqrt{xx}\) is:
A. x B. 1 C. 1 D. x E. \(\sqrt{x}\) if x <0 then x = x so this way we get \(\sqrt{(x)*(x)}\) i.e. \(\sqrt{x^2}\) i.e. x. As already said, x = x, Option A is correct. This is incorrect. If x < 0, then we get \(\sqrt{ (x)*(x)}\) i.e. \(\sqrt{x^2}\) HKD1710 has already provided a good explanation but you need to realize that when you assume x<0 , only the x portion of the expression \(\sqrt{(x)*x}\) will change to \(\sqrt{(x)*(x)}\) and NOT \(\sqrt{ (x)*(x)}\) You are writing x as x as well which is NOT correct, giving you an extra 1 than what is should be. 'x' will remain as such. Hope this helps.




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