Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.6 - Radical Equations - Exercise Set - Page 559: 11


$x=3$ or $x=1$

Work Step by Step

Subtract 1 on both sides of the equation to obtain: $x-1=\sqrt{2x-2}$ Square both sides of the equation to obtain: $(x-1)^2=2x-2$ Use the rule $(a-b)^2 = a^2 - 2ab+b^2$ where $a=x$ and $b=1$ to obtain: $\\x^2-2x+1=2x-2$ Move all terms to the left side of the equation. Note that when a term is moved to the other side of the equation, its sign changes to its opposite. $x^2-2x+1-2x+2=0 \\x^2-4x+3=0$ Factor the trinomial to obtain: $(x-3)(x-1)=0$ Equate each factor to zero then solve each equation to obtain: $x -3 = 0 \text{ or } x - 1 = 0 \\x = 3 \text{ or } x = 1$
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