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: 14

Answer

$x=\left\{ \dfrac{1-2\sqrt{5}}{2},\dfrac{1+2\sqrt{5}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 4r^2-4r-19=0 ,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ The quadratic equation above has $a= 4 , b= -4 , c= -19 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(4)(-19)}}{2(4)} \\\\ x=\dfrac{4\pm\sqrt{16+304}}{8} \\\\ x=\dfrac{4\pm\sqrt{320}}{8} .\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{4\pm\sqrt{64\cdot5}}{8} \\\\ x=\dfrac{4\pm\sqrt{(8)^2\cdot5}}{8} \\\\ x=\dfrac{4\pm8\sqrt{5}}{8} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel4(1)\pm\cancel4(2)\sqrt{5}}{\cancel4(2)} \\\\ x=\dfrac{1\pm2\sqrt{5}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{1-2\sqrt{5}}{2} \\\\\text{OR}\\\\ x=\dfrac{1+2\sqrt{5}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{1-2\sqrt{5}}{2},\dfrac{1+2\sqrt{5}}{2} \right\} .$
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