Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - Chapter Review - Page 320: 27

Answer

$$p=\pm 3\sqrt{3} -1$$

Work Step by Step

We first try to get the variable squared alone on onside of the equation. Then, we take the square root in order to solve: $$(p+1)^2 = 27 \\ p+1 =\pm\sqrt{27} \\ p = \pm 3\sqrt{3} -1$$
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