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a^4 - b^4

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justinpatellaro
 Post subject: a^4 - b^4
 Post Posted: Sun Oct 24, 2010 3:36 pm 
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Hello... My tutor gave me some homework and he is out of town for a week....

If a^4 - b^4 =65, whats a possible value of b?

I factored...
(a^2-b^2)(a^2+b^2)=65
(a^2-2ab+b^2)(a^2+2ab+b^2)=65

Not sure what to do from here... Can someone help?
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sahilkhurana06
 Post subject: Re: a^4 - b^4
 Post Posted: Mon Oct 25, 2010 1:34 am 
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I don't think its a GMAT question..Anyways,here's the answer..

There is only one way to solve this equation,by putting a=3 and b=2..
3^4-2^4 = 81-16=65..
So b can only take a value of 2
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justinpatellaro
 Post subject: Re: a^4 - b^4
 Post Posted: Mon Oct 25, 2010 1:57 pm 
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I appreciate the help...

Is the only way to solve by plugging in numbers? Because after looking at the problem, that is what I figured was b=2^4

Can we solve for b by factoring out the equation?
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giteshr
 Post subject: Re: a^4 - b^4
 Post Posted: Tue Oct 26, 2010 11:37 pm 
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Course Students


Posts: 19
justinpatellaro wrote:
Hello... My tutor gave me some homework and he is out of town for a week....

If a^4 - b^4 =65, whats a possible value of b?

I factored...
(a^2-b^2)(a^2+b^2)=65
(a^2-2ab+b^2)(a^2+2ab+b^2)=65

Not sure what to do from here... Can someone help?


(a^2-b^2)(a^2+b^2) = (a+b)(a-b)(a^2+b^2)
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mschwrtz
 Post subject: Re: a^4 - b^4
 Post Posted: Wed Oct 27, 2010 3:58 pm 
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Posts: 506
I assume that a and b are said to be integers, otherwise b could be any real number.

With that constraint, yes, you could check values. Simply take the fourth power of the first few integers, and look for a gap of 64:

1^4=1
2^4=16
3^4=81

and you're done. a could be 3 or -3, b could be 2 or -2. If you try a few more integers, you'll see that the gaps are all much too large:

4^4=256
5^4=625
6^4=1296
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atul.prasad
 Post subject: Re: a^4 - b^4
 Post Posted: Sat Nov 13, 2010 4:50 pm 
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Posts: 34
Hi, I guess this problem cannot be solved for a single value if it is not mentioned that a and b are integers:

So assuming that and factoring it out:

a^4 - b^4 can be written as : (a-b)(a+b)(a^2 + b^2)
now if this were to be equal to 65, then the number itself could be written as a product of 3 integers.
it follows that : 65 = 1 x 5 x 13 is the only way you can write it in terms of 3 integers
intuitively, you could say a-b = 1 and a+b = 5
so a = 3 and b = 2
and a^2 + b^2 = 13 validates our solution.

Hence b = 2
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jnelson0612
 Post subject: Re: a^4 - b^4
 Post Posted: Sun Nov 14, 2010 12:10 am 
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ManhattanGMAT Staff


Posts: 1857
Great work atul! I like your strategy. Both yours and Michael's work perfectly.

_________________
Jamie Nelson
ManhattanGMAT Instructor
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