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 178: 10


The solutions are $x=0$, $x=3$ and $x=-\dfrac{3}{2}\pm\dfrac{3\sqrt{3}}{2}i$

Work Step by Step

$3x^{4}=81x$ Take $81x$ to the left side of the equation: $3x^{4}-81x=0$ Take out common factor $3x$: $3x(x^{3}-27)=0$ Factor the binomial inside the parentheses: $3x(x-3)(x^{2}+3x+9)=0$ Set all three factors equal to $0$ and solve each individual equation: $3x=0$ $x=\dfrac{0}{3}$ $x=0$ $x-3=0$ $x=3$ $x^{2}+3x+9=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=1$, $b=3$ and $c=9$ Substitute the known values into the formula and evaluate: $x=\dfrac{-3\pm\sqrt{3^{2}-4(1)(9)}}{2(1)}=\dfrac{-3\pm\sqrt{9-36}}{2}=\dfrac{-3\pm\sqrt{-27}}{2}=...$ $...=\dfrac{-3\pm3\sqrt{3}i}{2}=-\dfrac{3}{2}\pm\dfrac{3\sqrt{3}}{2}i$ The solutions are $x=0$, $x=3$ and $x=-\dfrac{3}{2}\pm\dfrac{3\sqrt{3}}{2}i$
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