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: 9


The solutions are $x=0$, $x=2$ and $x=-1\pm\sqrt{3}i$

Work Step by Step

$2x^{4}=16x$ Take $16x$ to the left side of the equation: $2x^{4}-16x=0$ Take out common factor $2x$: $2x(x^{3}-8)=0$ Factor the binomial inside the parentheses: $2x(x-2)(x^{2}+2x+4)=0$ Set all three factors equal to $0$ and solve each individual equation separately: $2x=0$ $x=\dfrac{0}{2}$ $x=0$ $x-2=0$ $x=2$ $x^{2}+2x+4=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=1$, $b=2$ and $c=4$ Substitute the known values into the formula and evaluate: $x=\dfrac{-2\pm\sqrt{2^{2}-4(1)(4)}}{2(1)}=\dfrac{-2\pm\sqrt{4-16}}{2}=\dfrac{-2\pm\sqrt{-12}}{2}=...$ $...=\dfrac{-2\pm2\sqrt{3}i}{2}=-1\pm\sqrt{3}i$ The solutions are $x=0$, $x=2$ and $x=-1\pm\sqrt{3}i$
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