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: 63


$x=\sqrt{3}$ and $x=\sqrt{2}$

Work Step by Step

We solve by factoring: $\sqrt[4]{5x^{2}-6}=x$ $(\sqrt[4]{5x^{2}-6})^{4}=x^{4}$ $5x^{2}-6=x^{4}$ $x^{4}-5x^{2}+6=0$ $(x^2-3)(x^2-2)=0$ $(x^2-3)=0$ or $(x^2-2)=0$ $x^2=3$ or $x^2=2$ $x=\pm\sqrt{3}$ or $x=\pm\sqrt{2}$ However, the negative solutions $x=-\sqrt{3}$ and $x=-\sqrt{2}$ do not work in the original equation (because the 4th root on the left side of the equation is never negative), so we throw these solutions out. Thus, the only solution are: $x=\sqrt{3}$ and $x=\sqrt{2}$
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