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 632: 62



Work Step by Step

"4 times a number is decreased by 3" is written as $4x-3$. The equation can be written as: $(4x-3)^2=9$ Take the square root of each side. $\sqrt{(4x-3)^2}=\sqrt{9}$ Simplify. $4x-3=\pm3$ Add 3 to each side. $4x-3+3=\pm3+3$ Simplify $4x=3+3$ or $4x=-3+3$ $4x=6$ or $4x=0$ Divide each side by 4. $4x\div4=6\div4$ or $4x\div4=0\div4$ $x=1.5$ or $x=0$ State as a solution set. $x=\{0,1.5\}$
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