Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises - Page 512: 81


$m=\left\{ -2-3i,-2+3i \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $ m^2+4m+13=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} m^2+4m=-13 .\end{array} In the equation above, $b= 4 .$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then \begin{array}{l}\require{cancel} \left( \dfrac{4}{2} \right)^2 \\\\= \left( 2 \right)^2 \\\\= 4 .\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+4m+4=-13+4 \\\\ (m+2)^2=-9 .\end{array} Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} m+2=\pm\sqrt{-9} .\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+2=\pm\sqrt{-1}\cdot\sqrt{9} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} m+2=\pm i\sqrt{9} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} m+2=\pm i\sqrt{(3)^2} \\\\ m+2=\pm i(3) \\\\ m+2=\pm 3i .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} m=-2\pm 3i .\end{array} Hence, $ m=\left\{ -2-3i,-2+3i \right\} .$
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