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 - Page 352: 43


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

Work Step by Step

Using $(a+b)(c+d)=ac+ad+bc+bd$ or the Distributive Property, the given expression, $ (x-7)(x+1)=-16 ,$ is equivalent to \begin{array}{l}\require{cancel} x(x)+x(1)-7(x)-7(1)=-16 \\\\ x^2+x-7x-7=-16 \\\\ x^2+x-7x-7+16=0 \\\\ x^2-6x+9=0 .\end{array} Factoring the above equation, $ x^2-6x+9=0 ,$ results to \begin{array}{l}\require{cancel} 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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