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: 59

Answer

$x=\{-2,8\}$

Work Step by Step

"The difference between a number and 3" is written as $x-3$. The equation can be written as: $(x-3)^2=25$ Take the square root of each side. $\sqrt{(x-3)^2}=\sqrt{25}$ Simplify. $x-3=\pm5$ Add 3 to each side. $x-3+3=\pm5+3$ Simplify $x=5+3$ or $x=-5+3$ $x=8$ or $x=-2$ State as a solution set. $x=\{-2,8\}$
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