## Intermediate Algebra (12th Edition)

$x=\left\{ -4,6 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $x^2-2x-24=0 ,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} x^2-2x=24 .\end{array} In the equation above, $b= -2 .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{-2}{2} \right)^2 \\\\= \left( -1 \right)^2 \\\\= 1 .\end{array} Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to \begin{array}{l}\require{cancel} x^2-2x+1=24+1 \\\\ x^2-2x+1=25 .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} (x-1)^2=25 .\end{array} Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x-1=\pm\sqrt{25} \\\\ x-1=\pm\sqrt{(5)^2} \\\\ x-1=\pm5 \\\\ x=1\pm5 .\end{array} The solutions are \begin{array}{l}\require{cancel} x=1-5 \\\\ x=-4 \\\\\text{OR}\\\\ x=1+5 \\\\ x=6 .\end{array} Hence, $x=\left\{ -4,6 \right\} .$