Question 839474
The first thing we need to do is find the roots of {{{x^2-3x+1=0}}}.


You will notice that this polynomial will not factor evenly, so we will have to find the roots by using the quadratic formula: {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


Doing this will give us our two roots:  {{{((3+sqrt(5))/2)}}} and {{{((3-sqrt(5))/2)}}}


To find a new quadratic equation with roots that are one more than each of the roots of our original equation, we need to add 1 to each of the original roots:


{{{((3+sqrt(5))/2)+1}}} = {{{((5+sqrt(5))/2)}}} and


{{{((3-sqrt(5))/2)+1}}} = {{{((5-sqrt(5))/2)}}}


Next, we will put these roots into factored form:


{{{(x-((5+sqrt(5))/2))*(x-((5-sqrt(5))/2))}}}


Finally, we need to expand our factors by multiplying via the FOIL method, which will give us our final answer:


{{{x^2-5x+5=0}}}