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: 61



Work Step by Step

"Three times a number is increased by 2" is written as $3x+2$. The equation can be written as: $(3x+2)^2=49$ Take the square root of each side. $\sqrt{(3x+2)^2}=\sqrt{49}$ Simplify. $3x+2=\pm7$ Subtract 2 from each side. $3x+2-2=\pm7-2$ Simplify $3x=7-2$ or $3x=-7-2$ $3x=5$ or $3x=-9$ Divide each side by 3. $3x\div3=5\div3$ or $3x\div3=-9\div3$ $x=\frac{5}{3}$ or $x=-3$ State as a solution set. $x=\{-3,\frac{5}{3}\}$
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