Math questions from mba.com and GMAT Prep software
RonPurewal
Students
 
Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am
 

Re: If M = square root 4 + cube root 4 + fourth root of 4, then

by RonPurewal Fri Dec 12, 2014 5:20 am

nope. that's ... not a thing at all, actually. there's no way to simplify a sum of powers.

there's nothing wrong with having that idea in your head--from time to time, we all toss around ideas that turn out to be wrong--but it's very easy to disprove that idea with simple examples.
e.g., 1^2 + 1^3 + 1^4 is actually 3, but your method of figuring would give 1^9 = 1.
RonPurewal
Students
 
Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am
 

Re: If M = square root 4 + cube root 4 + fourth root of 4, then

by RonPurewal Fri Dec 12, 2014 5:22 am

more generally--if there were any way to simplify a sum of powers, then polynomials wouldn't exist!

you know, those things like x^2 + x + 6 that you spent so much time factoring in high school.
if you could simplify something like x^2 + x^1 into a single term, then those things wouldn't be things.
NicholasW85
Course Students
 
Posts: 1
Joined: Tue Sep 20, 2016 10:17 am
 

Re: If M = square root 4 + cube root 4 + fourth root of 4, then

by NicholasW85 Thu Dec 08, 2016 1:52 pm

I understand the estimation theory, but why couldnt you have removed a common base of 2 from all the terms in the problem. See below for how I thought of the problem:

sqrt4 = 2
cbrt4 = 2^(1/3)
fthrt4 = 2^(1/4)

Since they all have a 2 as the base then I figured I could re-write the equation as:

2*(1 + 1^1/3 + 1^1/4)

Calculating the numbers inside the parentheses is easy since they're all equal to 1, right? So you get:

2*(1+1+1) = 6 (greater than 4)

I may be missing something so please do let me know!

Thanks!
RonPurewal
Students
 
Posts: 19747
Joined: Tue Aug 14, 2007 8:23 am
 

Re: If M = square root 4 + cube root 4 + fourth root of 4, then

by RonPurewal Sun Dec 18, 2016 2:53 pm

that's definitely incorrect.
√4 is equal to 2, so, the cube root of 4 and the fourth root of 4 are clearly smaller than 2. therefore, the whole sum is most certainly less than 6.

the problem is that you're trying to factor out integers that you can't factor out.

for instance, √10 equals √2 times √5... but, what you're doing would be like trying to make √10 equal 2 times √5, or 5 times √2. nope. can't do that.