Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set: 36

Answer

$x=\left\{ -3 \right\}$

Work Step by Step

Factoring the given equation, $ 0=6x+x^2+9 ,$ results to \begin{array}{l}\require{cancel} 6x+x^2+9=0 \\\\ x^2+6x+9=0 \\\\ (x+3)^2=0 \\\\ (x+3)(x+3)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ (x+3)(x+3)=0 ,$ are \begin{array}{l}\require{cancel} x+3=0 \\\\ x=-3 ,\\\\\text{OR}\\\\ x+3=0 \\\\ x=-3 .\end{array} Hence, $ x=\left\{ -3 \right\} .$
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