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Dragonfinity · 2005 February 17th, 01:37 AM

Easy. Just set x^12 = x^2 then
x^12 - x^2 = 0
x^2*(x^10 - 1) = 0
x^2*(x^5 - 1)*(x^5 + 1) = 0
x = 0, 1, -1
So, we have (0,0), (1,1), and (-1,1).

2005 February 17th, 01:37 AM	#3 (permalink)
Dragonfinity Socratic realist Join Date: Feb 2005 Location: Georgia, USA Posts: 220	Re: Intersection Easy. Just set x^12 = x^2 then x^12 - x^2 = 0 x^2(x^10 - 1) = 0 x^2(x^5 - 1)*(x^5 + 1) = 0 x = 0, 1, -1 So, we have (0,0), (1,1), and (-1,1).