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 501: 29


The solutions are $-\sqrt3, -\sqrt2, \sqrt2, \text{ and } \sqrt3$.

Work Step by Step

The given equation can be written as: $(a^2)^2-5a^2+6=0$ Let $u=a^2$. Rewrite the equation above using $u$ to obtain: $u^2-5u+6=0$ Factor the trinomial to obtain: $(u-3)(u-2)=0$ Factor each factor with a variable to zero then solve each equation to obtain: \begin{array}{ccc} &u-3=0 &\text{or} &u-2=0 \\&u=3 &\text{or} &u=2 \end{array} Since $u=a^2$, then: \begin{array}{ccc} &u=3 &\text{or} &u=-\frac{5}{2} \\&a^2=3 &\text{or} &a^2=2 \\&(u=\pm\sqrt3 &\text{or} &u=\pm \sqrt2 \end{array} Therefore, the solutions are $-\sqrt3, -\sqrt2, \sqrt2, \text{ and } \sqrt3$.
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