College Algebra (6th Edition)

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

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 43


The solutions are $x=\pm1$ and $x=\pm\dfrac{4}{3}$

Work Step by Step

$9x^{4}=25x^{2}-16$ Take all terms to the left side of the equation: $9x^{4}-25x^{2}+16=0$ Let $u$ be equal to $x^{2}$ If $u=x^{2}$, then $u^{2}=x^{4}$ Rewrite the equation using the new variable $u$: $9u^{2}-25u+16=0$ Solve by factoring: $(u-1)(9u-16)=0$ Set both factors equal to $0$ and solve each individual equation for $u$: $u-1=0$ $u=1$ $9u-16=0$ $9u=16$ $u=\dfrac{16}{9}$ Substitute $u$ back to $x^{2}$ and solve for $x$: $x^{2}=1$ $\sqrt{x^{2}}=\pm\sqrt{1}$ $x=\pm1$ $x^{2}=\dfrac{16}{9}$ $\sqrt{x^{2}}=\pm\sqrt{\dfrac{16}{9}}$ $x=\pm\dfrac{4}{3}$ The solutions are $x=\pm1$ and $x=\pm\dfrac{4}{3}$
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