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: 7



Work Step by Step

Square both sides of the equation to obtain: $x^2=7x+8$ Move all terms on 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. $x^2-7x-8=0$ Factor the trinomial to obtain: $(x-8)(x+1) = 0$ Equate each factor to zero then solve each equation to obtain: $x-8 = 0 \text{ or } x + 1 = 0 \\x = 8 \text{ or } x = -1$ The principal root of a square root cannot be negative. This means that $x=-1$ is an extraneous root. Therefore, the solution is $x=8$.
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