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: 83


$r=\left\{ \dfrac{-2-2i\sqrt{2}}{3},\dfrac{-2+2i\sqrt{2}}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $ 3r^2+4r+4=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{3r^2+4r+4}{3}=\dfrac{0}{3} \\\\ r^2+\dfrac{4}{3}r+\dfrac{4}{3}=0 \\\\ r^2+\dfrac{4}{3}r=-\dfrac{4}{3} .\end{array} In the equation above, $b= \dfrac{4}{3} .$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then \begin{array}{l}\require{cancel} \left( \dfrac{\dfrac{4}{3}}{2} \right)^2 \\\\= \left( \dfrac{4}{3}\div2 \right)^2 \\\\= \left( \dfrac{4}{3}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( \dfrac{2}{3} \right)^2 \\\\= \dfrac{4}{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} r^2+\dfrac{4}{3}r+\dfrac{4}{9}=-\dfrac{4}{3}+\dfrac{4}{9} \\\\ \left( r+\dfrac{2}{3} \right)^2=-\dfrac{12}{9}+\dfrac{4}{9} \\\\ \left( r+\dfrac{2}{3} \right)^2=-\dfrac{8}{9} .\end{array} Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} r+\dfrac{2}{3}=\pm\sqrt{-\dfrac{8}{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} r+\dfrac{2}{3}=\pm\sqrt{-1}\cdot\sqrt{\dfrac{8}{9}} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} r+\dfrac{2}{3}=\pm i\sqrt{\dfrac{8}{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} r+\dfrac{2}{3}=\pm i\sqrt{\dfrac{4}{9}\cdot2} \\\\ r+\dfrac{2}{3}=\pm i\sqrt{\left(\dfrac{2}{3}\right)^2\cdot2} \\\\ r+\dfrac{2}{3}=\pm i\left(\dfrac{2}{3}\right)\sqrt{2} \\\\ r+\dfrac{2}{3}=\pm \dfrac{2}{3}i\sqrt{2} \\\\ r+\dfrac{2}{3}=\pm \dfrac{2i\sqrt{2}}{3} .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} r=-\dfrac{2}{3}\pm \dfrac{2i\sqrt{2}}{3} \\\\ r=\dfrac{-2\pm2i\sqrt{2}}{3} .\end{array} Hence, $ r=\left\{ \dfrac{-2-2i\sqrt{2}}{3},\dfrac{-2+2i\sqrt{2}}{3} \right\} .$
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