Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set - Page 502: 50


The solutions are $-3. -1, 1, \text{ and } 3$.

Work Step by Step

The given equation can be written as: $$(x^2)^2-10x^2+9=0$$ Let $u=x^2$. Rewrite the equation above using $u$ to obtain: $$u^2-10u+9=0$$ Factor the trinomial to obtain: $$(u-9)(u-1)=0$$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&u-9=0 &\text{or} &u-1=0 \\&u=9 &\text{or} &u=1 \end{array} Since $u=x^2$, then \begin{array}{ccc} \\&u=9 &\text{or} &u=1 \\&x^2=9 &\text{or} &x^2=1 \\&x=\pm\sqrt9 &\text{or} &x=\pm \sqrt1 \\&x=\pm3 &\text{or} &x=\pm 1 \end{array} Therefore, the solutions are $-3. -1, 1, \text{ and } 3$.
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