Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.3 - Equations Quadratic in Form - 8.3 Exercises - Page 530: 80

Answer

$\left\{1,\sqrt[3]{9}\right\}$

Work Step by Step

Using the properties of equality, the given equation, $ x^6-10x^3=-9 ,$ is equivalent to \begin{align*} x^6-10x^3+9&=0 .\end{align*} Using factoring of trinomials, the equation above is equivalent to \begin{align*} (x^3-9)(x^3-1)&=0 .\end{align*} Equating each factor to zero (Zero Product Property) and solving for the variable, then \begin{array}{l|r} x^3-9=0 & x^3-1=0 \\ x^3=9 & x^3=1 .\end{array} Taking the cube root of both sides, the equations above are equivalent to \begin{array}{l|r} x=\sqrt[3]{9} & x=\sqrt[3]{1} \\ & x=1 .\end{array} Hence, the solution set of the equation $ x^6-10x^3=-9 $ is $\left\{1,\sqrt[3]{9}\right\}$.
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