Algebra and Trigonometry 10th Edition

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

Chapter 1 - 1.5 - Complex Numbers - 1.5 Exercises - Page 120: 76

Answer

$\frac{1\pm i\sqrt {23} }{3}$

Work Step by Step

The quadratic formula says $x = \frac{ - b \pm \sqrt {b^2 - 4ac} }{2a}$ when $ax^2 + bx + c = 0$. We know that $i^2=-1$. If I multiply the equation by $2$ (this doesn't modify the solutions), the coefficients will be $9,-6,24$. Hence here $x=\frac{-(-6)\pm\sqrt{(-6)^2-4\cdot9\cdot24}}{2\cdot9}=\frac{6\pm\sqrt{36-864}}{18}=\frac{6\pm\sqrt{-828}}{18}=\frac{6\pm6i\sqrt {23}}{18}= \frac{1\pm i\sqrt {23} }{3}$
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.