College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 53


$x=\left\{ 3-\sqrt{2},3+\sqrt{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^2-6x=-7 ,$ express the equation in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} x^2-6x+7=0 .\end{array} In the equation above, $a= 1 ,$ $b= -6 ,$ and $c= 7 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(7)}}{2(1)} \\\\ x=\dfrac{6\pm\sqrt{36-28}}{2} \\\\ x=\dfrac{6\pm\sqrt{8}}{2} .\end{array} Extracting the perfect square factor of the radicand results to \begin{array}{l}\require{cancel} x=\dfrac{6\pm\sqrt{4\cdot2}}{2} \\\\ x=\dfrac{6\pm\sqrt{(2)^2\cdot2}}{2} \\\\ x=\dfrac{6\pm2\sqrt{2}}{2} \\\\ x=\dfrac{2(3\pm\sqrt{2})}{2} \\\\ x=\dfrac{\cancel2(3\pm\sqrt{2})}{\cancel2} \\\\ x=3\pm\sqrt{2} .\end{array} The solutions are $ x=\left\{ 3-\sqrt{2},3+\sqrt{2} \right\} .$
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