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: 22


$p=-\displaystyle \frac{1}{2}+\frac{\sqrt{15}}{2}i$ or $p=-\displaystyle \frac{1}{2}-\frac{\sqrt{15}}{2}i$

Work Step by Step

$ p^{2}+p+4=0\qquad$... solve with the Quadractic formula. $a=1,\ b=1,\ c=4$ $ p=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $1, \ a$ for $1$ and $c$ for $4$. $ p=\displaystyle \frac{-1\pm\sqrt{(1)^{2}-4\cdot(4)\cdot 1}}{2\cdot 1}\qquad$... simplify. $p=\displaystyle \frac{-1\pm\sqrt{1-16}}{2}$ $ p=\displaystyle \frac{-1\pm\sqrt{-15}}{2}\qquad$... write in terms of $i$. ($\sqrt{-1}=i$) $ p=\displaystyle \frac{-1\pm i\sqrt{15}}{2}\qquad$... the symbol $\pm$ indicates two solutions. $p=-\displaystyle \frac{1}{2}+\frac{\sqrt{15}}{2}i$ or $p=-\displaystyle \frac{1}{2}-\frac{\sqrt{15}}{2}i$
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