SOLUTION: x^4/3-5x^2/3+4=0
The answer the book gave is x=1 or x=+-8
Mustn't it be x=+-1 or x=+-8? When is it +- and when is it only 8.
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-> SOLUTION: x^4/3-5x^2/3+4=0
The answer the book gave is x=1 or x=+-8
Mustn't it be x=+-1 or x=+-8? When is it +- and when is it only 8.
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You can put this solution on YOUR website! x^4/3-5x^2/3+4=0
The answer the book gave is x=1 or x=+-8
Mustn't it be x=+-1 or x=+-8? When is it +- and when is it only 8.
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Your problem is a quadratic form:
[x^(2/3)]^2 - 5x^(2/3) + 4 = 0
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Let w = x^(2/3)
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w^2 - 5w + 4 = 0
Factor:
(w-4)(w-1) = 0
w = 4 or w = 1
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Solve for "x":
When x^(2/3) = 4
x = 4^(3/2)
x = 8
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When x^(2/3) = 1
x = 1^(3/2)
x = 1
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Your problem could be treated as a 4th degree equation:
x^4/3-5x^2/3+4=0
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[x^(1/3)]^4-5[x^(1/3)]^2+4 = 0
Let w = x^(1/3)
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w^4 - 5w^2 + 4 = 0
Factor:
(w^2-4)(w^2-1) = 0
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w = +-2 or w = +-1
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Solve for "x":
x^(1/3) = 2 then x = 8
x^(1/3) = -2 then x = -8
x^(1/3) = 1 then x = 1
x^(1/3) = -1 then x = -1
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Cheers,
Stan H.
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