Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Section 11.2 - Solving Quadratic Equations by Completing the Square - Exercise Set - Page 775: 20



Work Step by Step

$y^{2}-8=4y$ Take the $4y$ to the left side of the equation: $y^{2}-4y-8=0$ Use the quadratic formula to solve this equation. The formula is $y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this equation, $a=1$, $b=-4$ and $c=-8$ Substitute: $y=\dfrac{-(-4)\pm\sqrt{(-4)^{2}-4(1)(-8)}}{2(1)}=\dfrac{4\pm\sqrt{16+32}}{2}=...$ $...=\dfrac{4\pm\sqrt{48}}{2}=\dfrac{4\pm4\sqrt{3}}{2}=2\pm2\sqrt{3}$
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.