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: 31


$r=\left\{ 3-i\sqrt{5},3+i\sqrt{5} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ r^2-6r+14=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ The quadratic equation above has $a= 1 , b= -6 , c= 14 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} r=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(14)}}{2(1)} \\\\ r=\dfrac{6\pm\sqrt{36-56}}{2} \\\\ r=\dfrac{6\pm\sqrt{-20}}{2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} r=\dfrac{6\pm\sqrt{-1}\cdot\sqrt{20}}{2} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} r=\dfrac{6\pm i\sqrt{20}}{2} .\end{array} Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} r=\dfrac{6\pm i\sqrt{4\cdot5}}{2} \\\\ r=\dfrac{6\pm i\sqrt{(2)^2\cdot5}}{2} \\\\ r=\dfrac{6\pm i(2)\sqrt{5}}{2} \\\\ r=\dfrac{6\pm 2i\sqrt{5}}{2} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} r=\dfrac{\cancel2(3)\pm \cancel2(1)i\sqrt{5}}{\cancel2(1)} \\\\ r=3\pm i\sqrt{5} .\end{array} The solutions are \begin{array}{l}\require{cancel} r=3-i\sqrt{5} \\\\\text{OR}\\\\ r=3+i\sqrt{5} .\end{array} Hence, $ r=\left\{ 3-i\sqrt{5},3+i\sqrt{5} \right\} .$
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