## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 50

#### Answer

${{x}^{2}}-4x-6=0$

#### Work Step by Step

$2-\sqrt{10}\ \text{and 2}+\sqrt{10}$ It can be expressed as: $x=\text{2}+\sqrt{10}$ and $x=2-\sqrt{10}$ \begin{align} & x-\left( \text{2}+\sqrt{10} \right)=0 \\ & \left( x-2 \right)-\sqrt{10}=0 \end{align} Or, \begin{align} & x-\left( 2-\sqrt{10} \right)=0 \\ & \left( x-2 \right)+\sqrt{10}=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-2 \right)-\sqrt{10} \right)\left( \left( x-2 \right)+\sqrt{10} \right)=0 \\ & \left( x-2 \right)\left( x-2 \right)+\sqrt{10}\left( x-2 \right)-\sqrt{10}\left( x-2 \right)-{{\left( \sqrt{10} \right)}^{2}}=0 \end{align} Combine like terms: ${{\left( x-2 \right)}^{2}}-{{\left( \sqrt{10} \right)}^{2}}=0$ Apply the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ \begin{align} & {{x}^{2}}+4-4x-\left( 10 \right)=0 \\ & {{x}^{2}}-4x+4-10=0 \\ & {{x}^{2}}-4x-6=0 \end{align}

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