Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 24


$x=3+\sqrt{2}i$ or $x=3-\sqrt{2}i$

Work Step by Step

$ x^{2}+11=6x\qquad$...add $-6x$ to both sides. $ x^{2}+11-6x=0\qquad$... solve with the Quadractic formula. $a=1,\ b=11,\ c=-6$ $ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $-6,\ a$ for $1$ and $c$ for $11$. $ x=\displaystyle \frac{-(-6)\pm\sqrt{(-6)^{2}-4\cdot(11)\cdot 1}}{2\cdot 1}\qquad$... simplify. $x=\displaystyle \frac{6\pm\sqrt{36-44}}{2}$ $x=\displaystyle \frac{6\pm\sqrt{-8}}{2}$ $ x=\displaystyle \frac{6\pm\sqrt{-1\cdot 4\cdot 2}}{}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$) $x=\displaystyle \frac{6\pm 2\sqrt{2}i}{2}$ $ x=3\pm\sqrt{2}i\qquad$... the symbol $\pm$ indicates two solutions. $x=3+\sqrt{2}i$ or $x=3-\sqrt{2}i$
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.