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Topic: find x in ^(1/4) equation (Read 445 times) |
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gkwal
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find x in ^(1/4) equation
« on: Jul 16th, 2007, 9:39pm » |
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(x+27)^(1/4) + (55-x)^1/4 =4; x=?
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ThudnBlunder
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Re: find x in ^(1/4) equation
« Reply #1 on: Jul 17th, 2007, 9:04am » |
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x = 54 or -26
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Barukh
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Re: find x in ^(1/4) equation
« Reply #2 on: Jul 17th, 2007, 9:24am » |
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u + v = 4, u4 + v4 = 82.
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srn437
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Re: find x in ^(1/4) equation
« Reply #3 on: Sep 6th, 2007, 8:24pm » |
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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.
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mikedagr8
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Re: find x in ^(1/4) equation
« Reply #4 on: Sep 7th, 2007, 3:18am » |
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on Sep 6th, 2007, 8:24pm, 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. |
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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. 
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.
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| « Last Edit: Sep 7th, 2007, 4:04am by mikedagr8 » |
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srn437
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Re: find x in ^(1/4) equation
« Reply #5 on: Sep 8th, 2007, 8:12am » |
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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
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