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Re: Square root and Modulus [#permalink]
02 Sep 2010, 13:07

1

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Expert's post

mbafall2011 wrote:

udaymathapati wrote:

If x < 0, then \sqrt{-x} •|x|) is A. -x B. -1 C. 1 D. x E. \sqrt{x}

what is the source of this question. I havent seen any gmat question testing imaginary numbers

GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers. So you won't see any question involving imaginary numbers.

This question also does not involve imaginary numbers as expression under the square root is non-negative (actually it's positive): we have \sqrt{-x*|x|} --> as x<0 then -x=positive and |x|=positive, so \sqrt{-x*|x|}=\sqrt{positive*positive}=\sqrt{positive}.

Re: Square root and Modulus [#permalink]
08 Dec 2010, 11:34

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Expert's post

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 can not 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).

Re: Is the square root always positive? [#permalink]
01 Feb 2011, 05:23

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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 non-negative number is not always positive, it's never negative (always non-negative).

2. As x<0 then -x=-negative=positive so the answer is still positive.

For the complete solution refer to the above posts. _________________

Re: Square root and Modulus [#permalink]
30 Dec 2010, 14:20

Expert's post

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.

Re: Square root and Modulus [#permalink]
30 Dec 2010, 14:47

Expert's post

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?

I'm not sure understood your question.

What do you mean "change the value to -5"? 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.

Re: Square root and Modulus [#permalink]
31 Dec 2010, 08:06

Expert's post

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. _________________

Re: Square root and Modulus [#permalink]
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=-5 probably marijose is confusing the negative sign - with negative x .... \sqrt{-(x)*|x|} _________________

Re: Square root and Modulus [#permalink]
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 _________________

I'm the Dumbest of All !!

gmatclubot

Re: Square root and Modulus
[#permalink]
31 Dec 2010, 11:31

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