College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 101: 36


The solution is $\left\{0, \frac{4}{3}\right\}$.

Work Step by Step

Take the square root of both sides to obtain: $3z-2 = \pm \sqrt{4} \\3z-2 = \pm \sqrt{2^2} \\3z-2 = \pm 2$ Add 2 to both sides to obtain: $3z = 2 \pm2$ Divide 3 on both sides of the equation to obtain: $z=\dfrac{2 \pm 2}{3}$ Split the solutions to obtain: $z_1 = \dfrac{2+2}{3} =\dfrac{4}{3} \\z_2 = \dfrac{2-2}{3} = 0$ Thus, the solution is $\left\{0, \frac{4}{3}\right\}$.
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