Question 189450
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 = x]


Since this is a 2nd degree equation, we know that it must either have two roots or a single root with a multiplicity of 2.  So if <i>x</i> = 1 is a single root, then the factors of the polynomial must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 1)(x - 1)]


which is equal to


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 2x + 1]


and which is different than


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 = x]


Therefore, while one of the roots of 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 = x]


is <i>x</i> = 1, there must be a second root different from 1.


The proper method to solve this problem is to factor the polynomial obtained by putting the equation into standard quadratic form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2 - x = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x(x - 1) = 0]


hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>