College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 163: 44


$x=\left\{ \sqrt{2}-1,\sqrt{2}+1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \sqrt{2}x^2-4x+\sqrt{2}=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ In the equation above, $a= \sqrt{2} ,$ $b= -4 ,$ and $c= \sqrt{2} .$ 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{-(-4)\pm\sqrt{(-4)^2-4(\sqrt{2})(\sqrt{2})}}{2(\sqrt{2})} \\\\ x=\dfrac{4\pm\sqrt{16-8}}{2(\sqrt{2})} \\\\ x=\dfrac{4\pm\sqrt{8}}{2(\sqrt{2})} \\\\ x=\dfrac{4\pm\sqrt{4\cdot2}}{2(\sqrt{2})} \\\\ x=\dfrac{4\pm\sqrt{(2)^2\cdot2}}{2(\sqrt{2})} \\\\ x=\dfrac{4\pm2\sqrt{2}}{2(\sqrt{2})} \\\\ x=\dfrac{\cancel2^2\pm\cancel2^1\sqrt{2}}{\cancel2^1(\sqrt{2})} \\\\ x=\dfrac{2\pm\sqrt{2}}{\sqrt{2}} .\end{array} Rationalizing the denominator results to \begin{array}{l}\require{cancel} x=\dfrac{2\pm\sqrt{2}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} \\\\ x=\dfrac{2\sqrt{2}\pm(\sqrt{2})^2}{(\sqrt{2})^2} \\\\ x=\dfrac{2\sqrt{2}\pm2}{2} \\\\ x=\dfrac{\cancel2^1\sqrt{2}\pm\cancel2^1}{\cancel2^1} \\\\ x=\sqrt{2}\pm1 .\end{array} The solutions are $ x=\left\{ \sqrt{2}-1,\sqrt{2}+1 \right\} .$
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