Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 362: 86

Answer

The solution set is $\left\{1, 7\right\}$.

Work Step by Step

Subtract $10$ from both sides: to obtain: $$2(k-4)^2=18$$ Divide $2$ to both sides: \begin{align*} \frac{2(x-4)^2}{2}&=\frac{18}{2}\\\\ (x-4)^2&=9 \end{align*} Take the square root of both sides to obtain: \begin{align*} \sqrt{(x-4)^2}&=\pm\sqrt9\\ x-4&=\pm3\\ x&=4\pm3\\ x_1&=4-3=1\\ x_2&=4+3=7 \end{align*} Checking: \begin{align*} 2(1-4)^2+10&=28\\ 2(-3)^2+10&=28\\ 2(9)+10&=28\\ 18+10&=28\\ 28&\stackrel{\checkmark}=28 \end{align*} \begin{align*} 2(7-4)^2+10&=28\\ 2(3)^2+10&=28\\ 2(9)+10&=28\\ 18+10&=28\\ 28&\stackrel{\checkmark}=28 \end{align*} Thus, the solution set is $\left\{1, 7\right\}$.
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