Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 46


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (2x+1)(x-2)=-3 ,$ use the FOIL Method and convert the given equation in the form $ax^2+bx+c=0.$ Then express the equation in factored form. Equate each factor to zero using the Zero Product Property. Finally, solve each resulting equation. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 2x(x)+2x(-2)+1(x)+1(-2)=-3 \\\\ 2x^2-4x+x-2=-3 \\\\ 2x^2+(-4x+x)+(-2+3)=0 \\\\ 2x^2-3x+1=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 2(1)=2 $ and the value of $b$ is $ -3 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2x^2-x-2x+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} (2x^2-x)-(2x-1)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(2x-1)-(2x-1)=0 .\end{array} Factoring the $GCF= (2x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x-1)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} 2x-1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x-1=0 \\\\ 2x=1 \\\\ x=\dfrac{1}{2} \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, $ x=\left\{ \dfrac{1}{2},1 \right\} .$
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