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: 35

Answer

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

Work Step by Step

Take the square root of both sides to obtain: $2y+3 = \pm \sqrt{9} \\2y+3 = \pm \sqrt{3^2} \\2y+3 = \pm 3$ Subtract 3 to both sides to obtain: $2y = \pm 3 - 3 \\2y = -3 \pm 3$ Divide 2 on both sides of the equation to obtain: $y=\dfrac{-3 \pm 3}{2}$ Split the solutions to obtain: $y_1 = \dfrac{-3+3}{2} =0 \\y_2 = \dfrac{-3-3}{2} = -3$ Thus, the solution is $\left\{-3, 0\right\}$.
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