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: 17


$x=\left\{ 1-\sqrt{5},1+\sqrt{5} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ \dfrac{x^2}{4}-\dfrac{x}{2}=1 ,$ express first 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} \dfrac{x^2}{4}-\dfrac{x}{2}-1=0 \\\\ 4\left( \dfrac{x^2}{4}-\dfrac{x}{2}-1\right)=4(0) \\\\ x^2-2x-4=0 .\end{array} The quadratic equation above has $a= 1 , b= -2 , c= -4 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(1)(-4)}}{2(1)} \\\\ x=\dfrac{2\pm\sqrt{4+16}}{2} \\\\ x=\dfrac{2\pm\sqrt{20}}{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{2\pm\sqrt{4\cdot5}}{2} \\\\ x=\dfrac{2\pm\sqrt{(2)^2\cdot5}}{2} \\\\ x=\dfrac{2\pm2\sqrt{5}}{2} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel2(1)\pm\cancel2(1)\sqrt{5}}{\cancel2(1)} \\\\ x=1\pm\sqrt{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=1-\sqrt{5} \\\\\text{OR}\\\\ x=1+\sqrt{5} .\end{array} Hence, $ x=\left\{ 1-\sqrt{5},1+\sqrt{5} \right\} .$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.