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:

(1) a - b = √100 This means that a-b is +ve. For GMAT sqrt of something can not be -ve. So, a-b=10 ((a-b)^2)+4ab = (a+b)^2 = 100-36 = 64 (a+b) = 8 or -8 Not Enough. 2) Enough

(a +b) = sqrt (64) ---> Does GMAT considers -ve value of sqrt ?

If not, then answer could be D.

Hi, GMAT does not consider -ive value of square root , that is \(\sqrt{64}=8\).. But \((a + b)^2 = 64\) means a+b=8 or a+b=-8 and if it is given \((a +b) =(\sqrt{64}) = 8\).. these are two different things
_________________

(a +b) = sqrt (64) ---> Does GMAT considers -ve value of sqrt ?

If not, then answer could be D.

Here is the reason the square root is always positive but a square considers both positive and negative value.

A positive number has two square roots: a positive square root (also called principal square root) and a negative square root. They are depicted as \(\sqrt{x}\) and \(-\sqrt{x}\). So by convention, \(\sqrt{x}\) implies we are talking about only the positive square root. Hence when you are given \(\sqrt{x}\) in the question, it implies that this is the principal square root only.

On the other hand, \(x^2 = 4\) has two solutions: x = 2 or -2 since x could take either of these values.
_________________