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 113: 128

Answer

$x^2+6x+4$

Work Step by Step

Write the quadratic as the product of two linear factors and the leading coefficient: $a(x-r_1)(x-r_2)=0$, where $r_1$ and $r_2$ are the solutions. $a[x-(-3+\sqrt 5)][x-(-3-\sqrt 5)]=0~~$ (Make $a=1$ since it can be any real number except $0$) $(x+3-\sqrt 5)(x+3+\sqrt 5)=0$ $(x+3)^2-(\sqrt 5)^2=0$ $x^2+6x+9-5=0$ $x^2+6x+4$ (general form)
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.