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:

can someone explain how to do this type of questions in less time. If M = (4)^.5 + (4)^.3 + (4)^.25 , then the value of M is 1)less than 3 2) equal to 3 3)between 3 and 4 4)equal to 4 5)greater than 4

here \sqrt{4}=2 1<4^(1/3)<2 1<4^(1/4)<2

hence the sum, S;

4<S

P.S.: GO FOR THE FIXED MINIMUM VALUE THE ANSWER WILL BECOME EASY.

Hope this helps...!!
_________________

Practice Practice and practice...!!

If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight.

Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the [#permalink]

Show Tags

21 Apr 2012, 12:14

I was wondering why we are not considering the negative roots of 4 in this case. For instance, sq root of 4 would be 2 and -2...same for the 4th root... Any rule / trick that I might be missing here?

I was wondering why we are not considering the negative roots of 4 in this case. For instance, sq root of 4 would be 2 and -2...same for the 4th root... Any rule / trick that I might be missing here?

Look forward to the answer.

Cheers!

Welcome to GMAT Club. Below might help to clear your doubts.

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, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\). 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 \(\sqrt[3]{-64} =-4\).

I noted down this question, but I forgot the source and cannot retrieve the correct answer. Needless to say the question asked for the approximate value of the equation & whether it was between, or greater/smaller than a combination of numbers, i.e. +/- 3 & 3/4, 3 & 4 and so forth. The question is whether there is a shortcut for simplifying this or looking at it under a different perspective other than sheer number sense, Thanks and apologies for being vague.

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...