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 - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 48



Work Step by Step

$5-2i\ \text{and }5+2i$ It can be expressed as: $x=5+2i$ and $x=5-2i$ $\begin{align} & x-\left( 5+2i \right)=0 \\ & \left( x-5 \right)-2i=0 \end{align}$ Or, $\begin{align} & x-\left( 5-2i \right)=0 \\ & \left( x-5 \right)+2i=0 \\ \end{align}$ Apply the zero-product property $\left( \left( x-5 \right)-2i \right)\left( \left( x-5 \right)+2i \right)=0$ $\begin{align} & \left( \left( x-5 \right)-2i \right)\left( \left( x-5 \right)+2i \right)=0 \\ & \left( x-5 \right)\left( x-5 \right)+2i\left( x-5 \right)-2i\left( x-5 \right)-4{{i}^{2}}= \end{align}$ Combine like terms, Here, ${{i}^{2}}=1$ which is a complex entry. ${{\left( x-5 \right)}^{2}}-4\left( {{i}^{2}} \right)=0$ Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ $\begin{align} & {{x}^{2}}+25-10x-4\left( -1 \right)=0 \\ & {{x}^{2}}+25-10x+4=0 \\ & {{x}^{2}}-10x+29=0 \end{align}$
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