2006-11-20 11:17:02 · 10 answers · asked by Chloe D 2 in Science & Mathematics ➔ Mathematics
This problem is best solved in a manner identical to how all other polynomials are solved. So x^2=x^3? This is the same as x^3-x^2=0, so factoring this yields x times x times (x-1)=0 so the only possible solutions are x=0 and x=1.
2006-11-20 11:35:13 · answer #1 · answered by Sciencenut 7 · 0⤊ 0⤋
x can either be 0 or 1 because 0 to any power is 0 will result in 0. And 1 to any power will be 1. No other numbers have this property
2006-11-20 19:35:08 · answer #2 · answered by ImDownHere 2 · 0⤊ 0⤋
looks like 1 or 0.
COMPLETE PROOF:
x^3=x^2 move x^2 to the left
x^3-x^2=0 now we have a cubic equation reduciable to a quadratic
factor out x^2 now we have x^2(x-1)=0 using corollary if ab=0 then a or b needs to be zero or both
result x=1 or 0 (two identical real roots from x^2 =0)
hence 0,0,1 are the roots / you get 3 roots becuase its a 3rd degree equation.
2006-11-20 19:42:54 · answer #3 · answered by iceboxxman 2 · 0⤊ 0⤋
0 and 1 would be the only possibilities
2006-11-20 19:33:09 · answer #4 · answered by gregory_s19 3 · 0⤊ 0⤋
the only possible answers are 0 and 1
2006-11-20 19:32:37 · answer #5 · answered by Maverik42 3 · 1⤊ 0⤋
x^2= x^3
so transfer x^2 to the reight
x^3-x^2 =0
or x^2(x-1) = 0
so x =0 or 1
2006-11-20 19:34:51 · answer #6 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
If x^2= x^3 then
x^3-x^2 = 0
x^2(x-1) = 0
x = 0 or 1.
2006-11-20 20:04:54 · answer #7 · answered by steiner1745 7 · 0⤊ 0⤋
The only answers can be 0 and 1 because they are the only numbers that can be multiplied by themselves and equal themselves
2006-11-20 19:37:21 · answer #8 · answered by lavender tots 4 · 0⤊ 0⤋
the only answer is 1
2006-11-20 19:32:31 · answer #9 · answered by iberius 4 · 0⤊ 1⤋
that's easy. 1 or 0. (It took me forever to answer cause Yahoo! was giving me issues with my password............)
2006-11-20 19:33:03 · answer #10 · answered by frodobaggins115 4 · 0⤊ 0⤋