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 - Page 631: 34


$x=\left\{ -4,10 \right\}$

Work Step by Step

The left side of the given equation, $ x^2-6x+9=49 ,$ is a perfect square trinomial whose factored form is \begin{array}{l}\require{cancel} (x-3)^2=49 .\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{49} \\\\ x-3=\pm\sqrt{(7)^2} \\\\ x-3=\pm7 \\\\ x=3\pm7 .\end{array} Hence, $ x=\left\{ -4,10 \right\} .$
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