Question 173565
{{{x^2=2x-1}}} → {{{x^2-2x+1=0}}} → {{{(x-1)^2=0}}} → {{{x=1}}}, therefore {{{alpha=1}}}


(a) {{{alpha^3=1^3=1}}} and {{{3*alpha - 2 = 3*1 - 2 = 1}}}


(b) {{{alpha^4-alpha^2=1^4-1^2=1-1=0}}} and {{{2*alpha - 2= 2*1-2= 0}}}