Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 120

Answer

$x=0$ and $x=2$

Work Step by Step

$\sqrt[3]{4x^{2}-4x}=x$ Cube both sides: $(\sqrt[3]{4x^{2}-4x})^{3}=x^{3}$ $4x^{2}-4x=x^{3}$ Take all terms to the right: $0=x^{3}-4x^{2}+4x$ Reorganize: $x^{3}-4x^{2}+4x=0$ Take out common factor $x$: $x(x^{2}-4x+4)=0$ The trinomial inside the parentheses is a perfect square trinomial. Factor it: $x(x-2)^{2}=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x=0$ $(x-2)^{2}=0$ $\sqrt{(x-2)^{2}}=\sqrt{0}$ $x-2=0$ $x=2$ The solutions found are $x=0$ and $x=2$. The original equation is true for both of these values. The final answer is: $x=0$ and $x=2$
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