Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 43


$x=\left\{ -\dfrac{1}{4},-\dfrac{3}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 8x^2+14x+3=0 ,$ 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 factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 8(3)=24 $ and the value of $b$ is $ 14 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 2,12 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 8x^2+2x+12x+3=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} (8x^2+2x)+(12x+3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(4x+1)+3(4x+1)=0 .\end{array} Factoring the $GCF= (4x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4x+1)(2x+3)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} 4x+1=0 \\\\\text{OR}\\\\ 2x+3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 4x+1=0 \\\\ 4x=-1 \\\\ x=-\dfrac{1}{4} \\\\\text{OR}\\\\ 2x+3=0 \\\\ 2x=-3 \\\\ x=-\dfrac{3}{2} .\end{array} Hence, $ x=\left\{ -\dfrac{1}{4},-\dfrac{3}{2} \right\} .$
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