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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

Bunuel Can you please help here? According to my understanding there are two cases in the GMAT: Case 1 :

sqrt[(-5)*(-5)] = -5

Case 2 :

sqrt[(5)*(5)] = 5

Is my understanding correct?

Thanks in advance Ankit

No, it's totally wrong.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3; \(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Again, the point here is that since 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 \(\sqrt{x^2}=|x|\).

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

So in other words sq rt of (x-3)^2 is +ve or -ve x-3? Sorry, but i don't really get that. I thought when we have a square of a sq rt, the answer is just what's under the sq rt, without any power.

It's not sq rt of 9 to be +-3, it's sq rt of 9^2, which is 9.
Please correct me if i am wrong.

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.

the unknown x in sqrt(x^2) can be negative - the outcome will be positive ---> sqrt(-2^2) = 2 ---> x=-2

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.

No... ... In bold, we can only say : sqrt(x^2) = |x|

KillerSquirrel gave an exemple that proves it as well

I'm sorry; I just don't understand the logic behind this. I can arrive at the correct answer by plugging in but I cannot see how the answer is not x - 3 when solving algebraically. Obviously, I'm missing some basic concept. Can you explain? Thanks.

by definition |x| = sqrt(x^2)

since |x| have two scenarios -x or x and sqrt(x^2) have the same scenarios -x or x.

Whew! Finally it makes sense. Thanks KS.

wait a second: it is true that |x| can have -ve or +ve values i.e. lxl = x or -x.

But sqrt(x^2) has only one i.e x not -x.

No... ... In bold, we can only say : sqrt(x^2) = |x|

KillerSquirrel gave an exemple that proves it as well

Fig, I am still not convinced. I am in line with what Honghu and hobbit said here:

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...