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: 94

Answer

$x_{1}\approx-0.93$ $x_{2}\approx0.93$

Work Step by Step

The quadratic formula will be used, but before doing so substitution will be used:$x^2=u$ therefore: $u^2+\sqrt2u-2=0$ $u=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where a=1, b=$\sqrt2$, and c=-2 $u=\frac{-(\sqrt2)\pm\sqrt{(\sqrt2)^2-4(1)(-2)}}{2(1)}$ $u=\frac{-\sqrt2\pm\sqrt{2+8}}{2}$ $u=\frac{-\sqrt2\pm\sqrt{10}}{2}$ There are two parts: First: $x^2=\frac{-\sqrt2+\sqrt{10}}{2}$ $\sqrt{x^2}=\sqrt{\frac{-\sqrt2+\sqrt{10}}{2}}$ $x\approx\pm0.93$ Second: $x^2=\frac{-\sqrt2-\sqrt{10}}{2}$ $\sqrt{x^2}=\sqrt{\frac{-\sqrt2-\sqrt{10}}{2}}$ It is not possible to solve since there is a negative value inside the radical. Plugging the first values into the original equation confirms that they are the only real solutions.
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