Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.4 - Quadratic Equations and Applications - 1.4 Exercises - Page 110: 33


{-8, 4}

Work Step by Step

$x^2+4x-32 = 0$ $x^2+4x+4-36=0$ $(x+2)^2-6^2 = (x+2-6)(x+2+6)=(x-4)(x+8)$ (as $x^2-y^2=(x-y)(x+y)$ Hence either (x-4)=0 that gives x=4 or (x+8)=0 that gives x=-8
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.