Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 39

{$4-\sqrt {15},4+\sqrt {15}$}

Given: $x^2-8x+1=0$ This can be re-written as: $x^2-8x=-1$ We need to add $(-4)^2$ to complete the square. Thus, $x^2-8x+16=-1+16$ or, $(x-4)^2=15$ or, $(x-4)=\pm \sqrt {15}$ or, $x=4- \sqrt {15},4+\sqrt {15}$ Hence, solution set is $x=${$4-\sqrt {15},4+\sqrt {15}$}