Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-5 Completing the Square - Practice and Problem-Solving Exercises - Page 564: 24



Work Step by Step

$g^2+11g-468$ or, $g^2+11g=468$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=11$ Therefore, $b^2=4ac$ $\implies$ $c=\dfrac{b^2}{4a}$ Thus, $c=\dfrac{b^2}{4a}=\dfrac{(11)^2}{4}=\dfrac{121}{4}$ To complete the square, add $\dfrac{121}{4}$ on both sides. $g^2+11g+\dfrac{121}{4}=468+\dfrac{121}{4}$ $\implies (g+\dfrac{11}{2})^2=\dfrac{1993}{4}$ $\implies (g+\dfrac{11}{2})=22.32$ and $\implies (g+\dfrac{11}{2})=-22.32$ or, $g=16.82,-27.82$
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.