College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises: 18

Answer

$x=0$ $x=4-2\sqrt{3}$ $x=4+2\sqrt{3}$

Work Step by Step

We factor out $y^3$: $y^{5}-8y^{4}+4y^{3}=0$ $y^{3}(y^{2}-8y+4)=0$ $y^{3}=0$ or $(y^{2}-8y+4)=0$ Left solution: $y=0$ Right solution: use quadratic formula ($a=1, b=-8, c=4$): $y^{2}-8y+4=0$ $y=\displaystyle \frac{-(-8)\pm\sqrt{(-8)^{2}-4*1*4}}{2(1)}=\frac{8\pm\sqrt{48}}{2}=4\pm 2\sqrt{3}$ So the solutions are: $x=0$, $x=4-2\sqrt{3}$ and $x=4+2\sqrt{3}$
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