Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 40


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given eqution, $ x^2+2x=8 ,$ 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} x^2+2x-8=0 .\end{array} In the trinomial expression above, the value of $c$ is $ -8 $ and the value of $b$ is $ 2 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-8 \}, \{ 2,-4 \}, \\ \{ -1,8 \}, \{ -2,4 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -2,4 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x-2)(x+4)=0 .\end{array} Equating each factor to zero (Zero Product Property), the solutions to the equation above are \begin{array}{l}\require{cancel} x-2=0 \\\\\text{OR}\\\\ x+4=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-2=0 \\\\ x=2 \\\\\text{OR}\\\\ x+4=0 \\\\ x=-4 .\end{array} Hence, $ x=\left\{ -4,2 \right\} .$
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