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: 10

Answer

$x=12$

Work Step by Step

Square both sides of the equation to obtain: $2x+1=(x-7)^2$ Use the rule $(a-b)^2 = a^2 - 2ab+b^2$ where $a=x$ and $b=7$ to obtain: $\\2x+1=x^2-14x+49$ Move all terms to the right side of the equation. Note that when a term is moved to the other side of the equation, its sign changes to its opposite. $0=x^2-14x+49-2x-1 \\0=x^2-16x+48$ Factor the trinomial to obtain: $0=(x-4)(x-12)$ Equate each factor to zero then solve each equation to obtain: $x -4 = 0 \text{ or } x - 12 = 0 \\x = 4 \text{ or } x = 12$ Note that when $x=4$, $\sqrt{2(4)+1}= 4-7 \\\sqrt{8+1} = -3$, Since the principal square root cannot be negative, this means that $4$ is an extraneous solution. Thus, the solution is $x=12$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.