Question 684079
If {{{4+sqrt(5)}}} is a root, then so is {{{4-sqrt(5)}}}; 
...
x = {{{4+-sqrt(5)}}}
x - 4 = +/- {{{sqrt(5)}}}
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{{{(x-4)^2 = sqrt(5)^2}}}
{{{(x-4)^2 = 5}}}
...
{{{x^2-8x+16 = 5}}}
{{{highlight_green(x^2-8x+11=0)}}}
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