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: 29

Answer

$x=\left\{ \dfrac{3-i\sqrt{15}}{2},\dfrac{3+i\sqrt{15}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ x^2-3x+6=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ The quadratic equation above has $a= 1 , b= -3 , c= 6 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(6)}}{2(1)} \\\\ x=\dfrac{3\pm\sqrt{9-24}}{2} \\\\ x=\dfrac{3\pm\sqrt{-15}}{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} x=\dfrac{3\pm\sqrt{-1}\cdot\sqrt{15}}{2} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{3\pm i\sqrt{15}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{3-i\sqrt{15}}{2} \\\\\text{OR}\\\\ x=\dfrac{3+i\sqrt{15}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{3-i\sqrt{15}}{2},\dfrac{3+i\sqrt{15}}{2} \right\} .$
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