Question 301676: Solve: (x^2 - 5x +5)^(x^2 - 9x +20) = 1
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Solve: (x^2 - 5x +5)^(x^2 - 9x +20) = 1
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Either (x^2 - 5x +5) = 1 or -1, or
x^2 - 9x +20 = 0
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If (x^2 - 5x +5) = -1, then the exponent has to be an even integer.
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x^2 - 9x +20 = 0
(x-5)*(x-4) = 0
x = 5
x = 4
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(x^2 - 5x +5) = +1
x^2 - 5x + 4 = 0
x = 1
x = 4
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Try the -1
(x^2 - 5x +5) = -1
x^2 - 5x + 6 = 0
x = 2
x = 3
The x=2 will give an even exponent and is valid
The x=3 --> an even exponent too.
x = 2, 3, 4 or 5
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