Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises - Page 364: 32



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality and express the given equation, $ 9x^2=6x-1 ,$ in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Once in factored form, equate each factor to zero (Zero Product Property). Finally, solve each resulting equation. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} 9x^2-6x+1=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 9(1)=9 $ and the value of $b$ is $ -6 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -3,-3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 9x^2-3x -3x +1 =0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (9x^2-3x )-(3x -1)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x(3x-1 )-(3x -1)=0 .\end{array} Factoring the $GCF= (3x-1 ) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x-1 )(3x-1)=0 \\\\ (3x-1 )^2=0 .\end{array} Getting the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{array}{l}\require{cancel} 3x-1=\pm\sqrt{0} \\\\ 3x-1=0 \\\\ 3x=1 \\\\ x=\dfrac{1}{3} .\end{array} Hence, the solution is $ x=\dfrac{1}{3} .$
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