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: 34


$x=\left\{ \dfrac{1-i\sqrt{6}}{3},\dfrac{1+i\sqrt{6}}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 9x^2-6x=-7 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 9x^2-6x+7=0 .\end{array} The quadratic equation above has $a= 9 , b= -6 , c= 7 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(9)(7)}}{2(9)} \\\\ x=\dfrac{6\pm\sqrt{36-252}}{18} \\\\ x=\dfrac{6\pm\sqrt{-216}}{18} .\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{6\pm\sqrt{-1}\cdot\sqrt{216}}{18} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{6\pm i\sqrt{216}}{18} .\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} x=\dfrac{6\pm i\sqrt{36\cdot6}}{18} \\\\ x=\dfrac{6\pm i\sqrt{(6)^2\cdot6}}{18} \\\\ x=\dfrac{6\pm i(6)\sqrt{6}}{18} \\\\ x=\dfrac{6\pm 6i\sqrt{6}}{18} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel6(1)\pm \cancel6(1)i\sqrt{6}}{\cancel6(3)} \\\\ x=\dfrac{1\pm i\sqrt{6}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{1-i\sqrt{6}}{3} \\\\\text{OR}\\\\ x=\dfrac{1+i\sqrt{6}}{3} .\end{array} Hence, $ x=\left\{ \dfrac{1-i\sqrt{6}}{3},\dfrac{1+i\sqrt{6}}{3} \right\} .$
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