Question 154368
Since {{{x=8}}} is the only solution (and we're dealing with a quadratic), this means that the solution {{{x=8}}} occurs twice. So this means that the solutions are


{{{x=8}}} or {{{x=8}}}



{{{x-8=0}}} or {{{x-8=0}}} Subtract 8 from both sides for each equation



{{{(x-8)(x-8)=0}}} Combine the equations using the zero product property



{{{x^2-16x+64=0}}} FOIL



So the equation {{{y=x^2-16x+64}}} has the solution {{{x=8}}} (with a multiplicity of 2)