This question was one my math 10 review test

(x^2+5x+5)^(x^2-6x+5)=1

I think there are 5 answers to this but I'm not quite sure.

(x^2+5x+5)^(x^2-6x+5) = 1
Anything to the 0 power is 1. Therefore that exponent is equal to 0. Or 1 to the power of anything is 1. So the x^2+5x+5 can be equal to 1. Do both.

x^2-6x+5 = 0
(x-5)(x-1) = 0
Either x-5=0 or x-1=0
Therefore x=5 or x=1

x^2+5x+5 = 1
x^2+5x+5-1 = 0
x^2+5x+4 = 0
(x+4)(x+1) = 0
Either x+4=0 or x+1=0
Therefore x=-4 or x=-1

Answer: x = -4,-1,1,5

x^2+5x+5 = -1
x^2+5x+5+1 = 0
x^2+5x+6 = 0
(x+3)(x+2) = 0
Either x+3=0 or x+2=0
Therefore x=-3 or x=2
x=2 would make the exponent -3, so that doesn't work.
x=-3 would make the exponent 32, so that does work.
Yeah, you're absolutely right! I didn't even think of that!

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a^0 = 1
since (x^2+5x+5)^(x^2-6x+5)=1
so (x^2-6x+5)=0
x^2-5x-x+5 = 0
x(x-5)-(x-5) = 0
(x-5)(x-1) = 0
x=5 or x=1

it also might be be possible that the base itself is 1
so(x^2+5x+5)=1
x^2+5x+4=0
x^2+4x+x+4 = 0
x(x+4)+(x+4)=0
(x+4)(x+1)=0
x= -4 or x= -1

x^2 + 5x + 5 = 1 or x^2 - 6x + 5 = 0
x^2 + 5x + 4 = 0 or (x - 5)(x - 1) = 0
(x + 4)(x + 1) = 0 or (x - 5)(x - 1) = 0
x = -4 or x = -1 or x = 5 or x = 1