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 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set: 29

Answer

$x=\{-3,13\}$

Work Step by Step

Using $a^2+2ab+b^2=(a+b)^2$ or the factoring of perfect square trinomials, the given equation, $ x^2-10x+25=64 ,$ is equivalent to \begin{array}{l}\require{cancel} (x-5)^2=64 .\end{array} Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the equation, $ (x-5)^2=64 ,$ are \begin{array}{l}\require{cancel} x-5=\pm\sqrt{64} \\\\ x-5=\pm\sqrt{(8)^2} \\\\ x-5=\pm8 \\\\ x=5\pm8 \\\\ x=\{-3,13\} .\end{array}
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