We have A^B = 1
case 1: A is 1, B is any real number
x^2 - 5x + 5 = 1
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x=1; (B is 20)
x=4; (B is 2)
x = {1,4} are solutions
case 2: B is 0, A is any nonzero real number
note: A=0 when x= (5 ± √20)/2; not for any integers
x^2 - 11x + 30 = 0
(x-5)(x-6) = 0
x = {5,6} are also solutions
case 3: A is -1, B is an even integer
x^2 - 5x + 5 = -1
x^2 - 5x + 6 = 0
(x-2)(x-3) = 0
x=2; (B is 12 which is even)
x=3; (B is 6 which is even)
x = {2,3} are also solutions
Solution set: {1,2,3,4,5,6}
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Posted by Larry
on 2024-01-18 11:18:48 |
solution
