College Algebra (10th Edition)

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

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 95

Answer

$t_{1}\approx-1.85$ $t_{2}\approx0.17$

Work Step by Step

The quadratic formula will be used, but before doing so substitution will be used:$1+t=u$ therefore: $u^2\pi=\pi+u$ $\pi u^2-u-\pi=0$ $u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=\pi, b=-1, and\space c=-\pi$ $u=\frac{-(-1)\pm\sqrt{(-1)^2-4(\pi)(-\pi)}}{2(\pi)}$ $u=\frac{1\pm\sqrt{1+4\pi^2}}{2\pi}$ There are two solutions: First: $1+t=\frac{1+\sqrt{1+4\pi^2}}{2\pi}$ $t=\frac{1+\sqrt{1+4\pi^2}}{2\pi}-1$ $t\approx0.17$ Second: $1+t=\frac{1-\sqrt{1+4\pi^2}}{2\pi}$ $t=\frac{1-\sqrt{1+4\pi^2}}{2\pi}-1$ $t\approx-1.85$ Plugging these values into the original equation does confirm that they are real solutions.
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