Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test: 15

Answer

$x=\left\{ -2,-\dfrac{2}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given eqution, $ 3x^2+8x=-4 ,$ express the equation in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Next step is to equate each factor to zero (Zero Product Property). Finally, solve each equation. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} 3x^2+8x+4=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 3(4)=12 $ and the value of $b$ is $ 8 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 2,6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x^2+2x+6x+4=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} (3x^2+2x)+(6x+4)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(3x+2)+2(3x+2)=0 .\end{array} Factoring the $GCF= (3x+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3x+2)(x+2)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} 3x+2=0 \\\\\text{OR}\\\\ x+2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x+2=0 \\\\ 3x=-2 \\\\ x=-\dfrac{2}{3} \\\\\text{OR}\\\\ x+2=0 \\\\ x=-2 .\end{array} Hence, $ x=\left\{ -2,-\dfrac{2}{3} \right\} .$
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