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: 13

Answer

$x=7$ or $x=3$

Work Step by Step

Isolate the radical on the right side to obtain: $x=2\sqrt{x-3} + 3 \\x-3 = 2\sqrt{x-3}$ Square both sides of the equation to obtain: $(x-3)^2=2^2(x-3)$ Use the rule $(a-b)^2 = a^2 - 2ab+b^2$ where $a=x$ and $b=3$ to obtain: $\\x^2-6x+9=4(x-3) \\x^2-6x+9=4x-12$ 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. $x^2-6x+9-4x+12=0 \\x^2-10x+21=0$ Factor the trinomial to obtain: $(x-7)(x-3)=0$ Equate each factor to zero then solve each equation to obtain: $x-7 = 0 \text{ or } x -3 = 0 \\x = 7 \text{ or } x = 3$
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