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 Fifth root (cubic) = Cube root (quintic) (Posted on 2009-04-05)
Find all possible real number(s) N that satisfy this equation:

5√(N3 + 2N) = 3√(N5 - 2N)

where, the common value of both sides of the equation is positive.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Oops! Solution by reading the problem. | Comment 4 of 5 |
(In reply to re: Oops! Solution by reading the problem. by Daniel)

Daniel:

Hmm.  Right you are!  I just substituted in N = +/- i, and I see that you are correct.  They are both solutions of the equation.  I wonder where I went wrong.

Also, once again,

5√(N3 + 2N) = 3√(N5 - 2N) = N

I just worked out why.

If  3√(N5 - 2N) = N
then (N5 - 2N) = N3 by cubing each side
then  N5 = N3 + 2N  by rearranging terms
and  N = 5√(N3 + 2N) by taking the fifth root of each side
so 5√(N3 + 2N) = 3√(N5 - 2N) = N

I also just noticed that if
If  3√(N5 - 2N) = N
then (N5 - 2N) = N3
then (N5 - N3 - 2N) =   0
and N ( N^2 + 1)(N^2 - 2) = 0
and N is 0 or sqrt(2) or -sqrt(2) or i or -i

I am not claiming these are the only solutions to the original equation

 Posted by Steve Herman on 2009-04-05 23:21:26

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