## Intermediate Algebra (12th Edition)

$m=\left\{ -3-i\sqrt{3},-3+i\sqrt{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $-m^2-6m-12=0 ,$ use the properties of equality to express the given equation in the form $x^2+bx=c.$ Then complete the square by adding $\left(\dfrac{b}{2} \right)^2$ to both sides. Factor the left side then take the square root (Square Root Property) of both sides. Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical and 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} \dfrac{-m^2-6m-12}{-1}=\dfrac{0}{-1} \\\\ m^2+6m+12=0 \\\\ m^2+6m=-12 .\end{array} In the equation above, $b= 6 .$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then \begin{array}{l}\require{cancel} \left( \dfrac{6}{2} \right)^2 \\\\= \left( 3 \right)^2 \\\\= 9 .\end{array} Adding $\left(\dfrac{b}{2} \right)^2$ to both sides of the equation above to complete the square, the equation becomes \begin{array}{l}\require{cancel} m^2+6m+9=-12+9 \\\\ (m+3)^2=-3 .\end{array} Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} m+3=\pm\sqrt{-3} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} m+3=\pm\sqrt{-1}\cdot\sqrt{3} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} m+3=\pm i\sqrt{3} .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} m=-3\pm i\sqrt{3} .\end{array} Hence, $m=\left\{ -3-i\sqrt{3},-3+i\sqrt{3} \right\} .$