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        riddles >> medium >> find x in  ^(1/4)  equation
        
(Message started by: gkwal on Jul 16^th, 2007, 9:39pm)

Title: find x in ^(1/4) equation
Post by gkwal on Jul 16^th, 2007, 9:39pm

(x+27)^(1/4) + (55-x)^1/4 =4; x=?

Title: Re: find x in ^(1/4) equation
Post by ThudanBlunder on Jul 17^th, 2007, 9:04am

[hide]x = 54 or -26[/hide]

Title: Re: find x in ^(1/4) equation
Post by Barukh on Jul 17^th, 2007, 9:24am

[hide]u + v = 4, u⁴ + v⁴ = 82[/hide].

Title: Re: find x in ^(1/4) equation
Post by srn347 on Sep 6^th, 2007, 8:24pm

V=1 u=3 or vice verca. Now here's one for you people to answer. (1-x)^1/3+(x-3)^1/3=1

Here's a hint, cubing it leads to extraneous solutions like x=2.

Title: Re: find x in ^(1/4) equation
Post by mikedagr8 on Sep 7^th, 2007, 3:18am

on 09/06/07 at 20:24:38, srn347 wrote:

V=1 u=3 or vice verca. Now here's one for you people to answer. (1-x)^1/3+(x-3)^1/3=1

Here's a hint, cubing it leads to extraneous solutions like x=2.

You need to spread it out more. Do you mean (1-x)^1/3 + (x-3)^1/3 = 1? Becuase if not, what do you mean, I want to get this correct. :P

It doesn't seem to be an answer. I seem to get 4=1 when I cube it. If I graph it won't intersect. I've also been able to make 4-x = x. This is not possible to do, if it is using the equation I stated above.

Title: Re: find x in ^(1/4) equation
Post by srn347 on Sep 8^th, 2007, 8:12am

It is what I mean. If you cube it , simplify, cube again, simplify agai, and apply quadratics, you get x=2, which doesn't work for the real cube root, but perhaps for the complex. http://en.wikipedia.org/wiki/Invalid_proof