Question 120471
{{{sqrt(x-1)=x-3}}}
{{{x-1=(x-3)^2}}}  (Square both sides)
{{{x-1=x^2-6x+9}}} (Expand binomial)
{{{0=x^2-7x+10}}}  (subtract (x-1) from both sides)
{{{0=(x-5)(x-2)}}} (factor right side)

giving the following two equations
{{{x-5=0}}} or {{{x-2=0}}}
which implies {{{x=5}}} or {{{x=2}}},

but {{{x=2}}} gives {{{1=-1}}} when substituted into the original equation, thus we exclude this solution, leaving {{{x=5}}} as the solution to this problem.