Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises - Page 519: 11


$x=\left\{ 5-\sqrt{7},5+\sqrt{7} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ x^2+18=10x ,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x^2-10x+18=0 .\end{array} The quadratic equation above has $a= 1 , b= -10 , c= 18 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4(1)(18)}}{2(1)} \\\\ x=\dfrac{10\pm\sqrt{100-72}}{2} \\\\ x=\dfrac{10\pm\sqrt{28}}{2} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} x=\dfrac{10\pm\sqrt{4\cdot7}}{2} \\\\ x=\dfrac{10\pm\sqrt{(2)^2\cdot7}}{2} \\\\ x=\dfrac{10\pm2\sqrt{7}}{2} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel2(5)\pm\cancel2(1)\sqrt{7}}{\cancel2(1)} \\\\ x=5\pm\sqrt{7} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=5-\sqrt{7} \\\\\text{OR}\\\\ x=5+\sqrt{7} .\end{array} Hence, $ x=\left\{ 5-\sqrt{7},5+\sqrt{7} \right\} .$
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