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Post subject: If M = square root 4 + cube root 4 + fourth root of 4, then  Posted: Wed Jun 27, 2007 7:13 pm |
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If M = square root 4 + cube root 4 + fourth root of 4, then the value of M is:
a) less than 3
b) equal to 3
c) between 3 and 4
d) equal to 4
e) greater than 4 - answer
Clearly if I had a calculator this would be easy. Am I just supposed to have memorized these numbers or is there a logical way to deal with this problem? Thanks!
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Jeff
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Post subject: Root problem Posted: Thu Jun 28, 2007 9:31 am |
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This is an estimation problem, you don't need to know precise values for the cube and fourth root of 4 - you just need to realize that they are each between 1 and 2. Here's why: The square root of 4 is 2. So clearly the cube and fourth root must be smaller than the square root, i.e. less than 2. However they must be greater than 1 - positive numbers less than 1 (say .75) get smaller as they are multiplied together and you need to "build-up" to 4.
So then you're left with sqrrt(4)+cubert(4)+4throot(4) = 2 + something between 1 and 2+ somthing between 1 and 2 so it must be greater than 4.
Jeff
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Guest
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Post subject: Posted: Thu Jun 28, 2007 10:55 am |
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Seriously, Jeff, you make this all sound so easy. :-)
Thanks. Your online explanations for this problem - as well as everyone else's problems - have been tremendously helpful.
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steph
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Post subject: ditto :-) Posted: Sat Sep 13, 2008 1:01 pm |
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RonPurewal
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Post subject: Posted: Thu Oct 09, 2008 7:38 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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well played.
there are a couple of other threads on this problem floating around, but they're hard to search (not a lot of keywords in this problem statement).
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Guest
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Post subject: Posted: Wed Dec 03, 2008 3:43 pm |
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sqrt (4) can be minus two as well. Also fourth root of 4 can also be negative. In which case the equation cannot be greater than 4. Hence it cannot be concretely said that it will be greater than 4.
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RonPurewal
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Post subject: Posted: Thu Dec 04, 2008 4:08 am |
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| ManhattanGMAT Staff |
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Anonymous wrote: sqrt (4) can be minus two as well. wrong. Quote: Also fourth root of 4 can also be negative.
also wrong.
here's the deal: you're talking about two different things.
when you have an OPERATOR SYMBOL, such as "√", that symbol will produce exactly ONE value.
for instance, √9 is just 3.
|-5| is just 5.
and so on.
notice, in these cases, that you're merely simplifying expressions; you are NOT solving EQUATIONS. in other words, there's no "=" sign in the middle of what you're dealing with; you're not finding the solutions of anything. just simplifying.
when you solve an EQUATION, there may well be more than one solution.
if you say x^2 = 9, there are two solutions: x = 3 and x = -3.
notice that i can say this without speaking of the "√" symbol at all!
indeed, the "√" symbol is DEFINED to refer only to the positive root. it is NOT defined to refer to either of the 2 solutions to the above equation.
same goes for 4th root, 6th root, and all other even roots.
hth.
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priti2000
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Post subject: Re: If M = square root 4 + cube root 4 + fourth root of 4, then Posted: Sat Nov 27, 2010 8:30 pm |
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I represented the roots as fractions and then factored out, so:
sqrt(4) + cube root(4) + fourth root (4) =
2 + 4^1/3 + 4^1/4 =
3 + 4^1/3
which is greater than 4
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atul.prasad
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Post subject: Re: If M = square root 4 + cube root 4 + fourth root of 4, then Posted: Sun Nov 28, 2010 7:46 am |
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priti2000 wrote: I represented the roots as fractions and then factored out, so:
sqrt(4) + cube root(4) + fourth root (4) =
2 + 4^1/3 + 4^1/4 =
3 + 4^1/3
which is greater than 4 What did you factor out? One key thing to note is that nth root of any positive number greater than 1, HAS to be greater than 1. this is because nth root of 1 is 1, hence the nth root of any number > 1 has to be > 1 as an example, lets prove by contradiction=> let x = 5^(1/29) or x^29 = 5.. now if x is < 1 , x^y WILL be < 1 for all values of y>=1 Hence for x^29 to be equal to 5, x cannot be < 1 Hence it follows that for our question, we have sqrt(4) + cube rt(4) + 4th root(4) 2+ (a number>1) + (a number > 1) = a number greater than 4
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jnelson0612
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Post subject: Re: If M = square root 4 + cube root 4 + fourth root of 4, then Posted: Sun Nov 28, 2010 2:41 pm |
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| ManhattanGMAT Staff |
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Posts: 1857
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Nice commentary everyone--thank you!
_________________
Jamie Nelson
ManhattanGMAT Instructor
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aseem83
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Post subject: Re: If M = square root 4 + cube root 4 + fourth root of 4, then Posted: Sat Oct 15, 2011 4:17 pm |
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Guys, just to add on here:
If 4(or any +ve number) raise to power 0 is = 1; then obviously any other root level(1/3,1/4,1/545) would be greater than, although by a miniscule, 1. Hence this value would always be more than 4!!
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RonPurewal
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Post subject: Re: If M = square root 4 + cube root 4 + fourth root of 4, then Posted: Sun Oct 16, 2011 2:26 am |
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| ManhattanGMAT Staff |
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aseem83 wrote: Guys, just to add on here:
If 4(or any +ve number) raise to power 0 is = 1; then obviously any other root level(1/3,1/4,1/545) would be greater than, although by a miniscule, 1. Hence this value would always be more than 4!! yes, this is another good way to remember that 4^fraction, or (any other number greater than 1)^fraction, is greater than 1. it's also a good way to remember that, say, (1/2)^fraction is less than 1.
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