College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Summary, Review, and Test - Cumulative Review Exercises (Chapters 1-8) - Page 792: 13

Answer

$$x = \frac{97\frac{+}{} \sqrt {193}}{72}$$

Work Step by Step

$$x^{\frac{1}{2}} - 6x^{\frac{1}{4}} + 8 = 0$$ We can re-write this equation in the following manner: $(x^{\frac{1}{2}}) - 6(x^{\frac{1}{2}})^{2} + 8 = 0$. Substituting the term $x^{\frac{1}{2}}$for an arbitrary variable $y$, we can now express the equation as: $$y - 6y^{2} + 8 = 0$$ And solve using the quadratic formula $\frac{-b\frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$ $$ y = \frac{-(1)\frac{+}{} \sqrt {(1)^{2} - 4(-6)(8)}}{2(-6)}$$ $$ y = \frac{-1\frac{+}{} \sqrt {1 + 192}}{-12}$$ $$ y = \frac{-1\frac{+}{} \sqrt {193}}{-12}$$ Substituting $y$ to its original value: $$(x^{\frac{1}{2}}) = \frac{-1\frac{+}{} \sqrt {193}}{-12}$$ $$(x^{\frac{1}{2}})^{2} = (\frac{-1\frac{+}{} \sqrt {193}}{-12})^{2}$$ $$x = \frac{1 + 193\frac{+}{} 2\sqrt {193}}{144}$$ $$x = \frac{194\frac{+}{} 2\sqrt {193}}{144}$$ $$x = \frac{97\frac{+}{} \sqrt {193}}{72}$$
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