Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set: 33

Answer

$x=\left\{ -3,9 \right\}$

Work Step by Step

The left side of the given equation, $ x^2-6x+9=36 ,$ is a perfect square trinomial whose factored form is \begin{array}{l}\require{cancel} (x-3)^2=36 .\end{array} Taking the square root of both sides (the Square Root Property), the solutions to the given equation are \begin{array}{l}\require{cancel} x-3=\pm\sqrt{36} \\\\ x-3=\pm\sqrt{(6)^2} \\\\ x-3=\pm6 \\\\ x=3\pm6 .\end{array} Hence, $ x=\left\{ -3,9 \right\} .$
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