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: 66


$x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 3r^2+2r-2=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} \dfrac{3r^2+2r-2}{3}=\dfrac{0}{3} \\\\ r^2+\dfrac{2}{3}r-\dfrac{2}{3}=0 \\\\ r^2+\dfrac{2}{3}r=\dfrac{2}{3} .\end{array} In the equation above, $b= \dfrac{2}{3} .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{\dfrac{2}{3}}{2} \right)^2 \\\\= \left( \dfrac{2}{3}\div{2} \right)^2 \\\\= \left( \dfrac{2}{3}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( \dfrac{1}{3} \right)^2 \\\\= \dfrac{1}{9} .\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} r^2+\dfrac{2}{3}r+\dfrac{1}{9}=\dfrac{2}{3}+\dfrac{1}{9} \\\\ r^2+\dfrac{2}{3}r+\dfrac{1}{9}=\dfrac{6}{9}+\dfrac{1}{9} \\\\ r^2+\dfrac{2}{3}r+\dfrac{1}{9}=\dfrac{7}{9} .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( r+\dfrac{1}{3} \right)^2=\dfrac{7}{9} .\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} r+\dfrac{1}{3}=\pm\sqrt{\dfrac{7}{9}} \\\\ r+\dfrac{1}{3}=\pm\sqrt{\dfrac{1}{9}\cdot7} \\\\ r+\dfrac{1}{3}=\pm\dfrac{1}{3}\sqrt{7} \\\\ r+\dfrac{1}{3}=\pm\dfrac{\sqrt{7}}{3} \\\\ r=-\dfrac{1}{3}\pm\dfrac{\sqrt{7}}{3} \\\\ r=\dfrac{-1\pm\sqrt{7}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} r=\dfrac{-1-\sqrt{7}}{3} \\\\\text{OR}\\\\ r=\dfrac{-1+\sqrt{7}}{3} .\end{array} Hence, $ x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\} .$
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