Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 non-negative 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 non-negative 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.

I'm having trouble understanding how they get to the answer, though it should be very simple:

If x<0, then \sqrt{-x|x|} 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..

_________________

Did you find this post helpful?... Please let me know through the Kudos button.

I'm having trouble understanding how they get to the answer, though it should be very simple:

If x<0, then \sqrt{-x|x|} 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.

are you sure the question is correct . I remember seeing the question as sqrt(-x/|x|) In that case ans is -1. but the problem you have stated, the ans should be x.

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 non-negative 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 non-negative 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?

\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 non-negative 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 non-negative 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.