## Intermediate Algebra for College Students (7th Edition)

The solutions are $3$ and $-9$, and the solution set is {$3, -9$}.
$x^2 - 6x + 9 = 36$ Subtract $9$ from both sides to isolate the binomial $x^2 - 6x$. $x^2 - 6x + 9 -9 = 36-9$ $x^2 - 6x = 27$ The coefficient of the x-term is $-6$. Half of $-6$ is $-3$, and $-3^2 = 9$. Thus, add $9$ to both sides of the equation to complete the square. $x^2 - 6x + 9= 27 +9$ $x^2 + 6x +9 = 36$ $(x-3)^2 = 36$ $x -3 =\sqrt36$ or $x -3 =-\sqrt36$ $x = -3+6$ or $x = -3-6$ The solutions are $3$ and $-9$, and the solution set is {$3, -9$}.