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: 49



Work Step by Step

$3-\sqrt{14}\ \text{and }3+\sqrt{14}$ It can be expressed as, $x=3+\sqrt{14}$ and $x=3-\sqrt{14}$ $\begin{align} & x-\left( 3+\sqrt{14} \right)=0 \\ & \left( x-3 \right)-\sqrt{14}=0 \end{align}$ Or, $\begin{align} & x-\left( 3-\sqrt{14} \right)=0 \\ & \left( x-3 \right)+\sqrt{14}=0 \\ \end{align}$ Apply the zero-product property $\left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0$ $\begin{align} & \left( \left( x-3 \right)-\sqrt{14} \right)\left( \left( x-3 \right)+\sqrt{14} \right)=0 \\ & \left( x-3 \right)\left( x-3 \right)+\sqrt{14}\left( x-3 \right)-\sqrt{14}\left( x-3 \right)-{{\left( \sqrt{14} \right)}^{2}}=0 \end{align}$ Combine like terms: Here ${{i}^{2}}=1$ which is a complex entry. ${{\left( x-3 \right)}^{2}}-{{\left( \sqrt{14} \right)}^{2}}=0$ Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ $\begin{align} & {{x}^{2}}+9-6x-\left( 14 \right)=0 \\ & {{x}^{2}}-6x+9-14=0 \\ & {{x}^{2}}-6x-5=0 \end{align}$
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