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: 40


The solutions are $x=2$ and $x=1$

Work Step by Step

$(x^{2}-3x+3)^{3/2}-1=0$ Take $1$ to the right side of the equation: $(x^{2}-3x+3)^{3/2}=1$ Raise both sides to $\dfrac{2}{3}$: $[(x^{2}-3x+3)^{3/2}]^{2/3}=1^{2/3}$ $x^{2}-3x+3=1$ Take $1$ to the left side and simplify: $x^{2}-3x+3-1=0$ $x^{2}-3x+2=0$ Solve by factoring: $(x-2)(x-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x-1=0$ $x=1$ The solutions are $x=2$ and $x=1$
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