Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.3 Complex Numbers - 1.3 Exercises - Page 113: 107


$-3+4i$ satisfies the equation.

Work Step by Step

$ x^{2}+6x+25=0\qquad$...substitute $x$ for $-3+4i$. $(-3+4i)^{2}+6(-3+4i)+25=0\qquad$...Apply the Perfect Square Formula: $(a+b)^{2}=a^{2}+2ab+b^{2},\ a=-3,\ b=4i$ $(-3)^{2}+2(-3)\cdot 4i+(4i)^{2}+6(-3+4i)+25=0$ $ 9-24i+16i^{2}+6(-3+4i)+25=0\qquad$...Apply imaginary number rule: $\quad i^{2}=-1$ $ 9-24i-16+6(-3+4i)+25=0\qquad$...Apply the distributive property: $\quad a(b-c)=ab-ac$ $ 9-24i-16+6(-3)+6(4i)+25=0\qquad$...Simplify. $-7-24i-18+24i+25=0\qquad$...add like terms. $0=0$
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