## College Algebra (10th Edition)

The solution set is $\left\{-\dfrac{8}{5}\right\}$.
Square both sides to obtain: $x^2-x-4=(x+2)^2$ Use the rule $(a+b)^2 = a^2+2ab+b^2$ to obtain: $x^2-x-4=x^2+2(x)(2) + 2^2 \\x^2-x-4=x^2+4x+4$ Put all terms on the left side of the equation. and then combine like terms. Note that when a term is transferred to the other side of an equation, its sign/operation changes to its opposite. $x^2-x-4-x^2-4x-4=0 \\(x^2-x^2)+(-x-4x) + (-4-4) = 0 \\-5x+(-8)=0 \\-5x-8=0$ Add $5x$ on both sides of the equation to obtain: $-8 = 5x$ Divide 5 on both sides of the equation to obtain: $\dfrac{-8}{5} = x$ Thus, the solution set is $\left\{-\dfrac{8}{5}\right\}$.