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: 33


$x=\left\{ \dfrac{1-i\sqrt{6}}{2},\dfrac{1+i\sqrt{6}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 4x^2-4x=-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} 4x^2-4x+7=0 .\end{array} The quadratic equation above has $a= 4 , b= -4 , c= 7 .$ 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)(7)}}{2(4)} \\\\ x=\dfrac{4\pm\sqrt{16-112}}{8} \\\\ x=\dfrac{4\pm\sqrt{-96}}{8} .\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{4\pm\sqrt{-1}\cdot\sqrt{96}}{8} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{4\pm i\sqrt{96}}{8} .\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{4\pm i\sqrt{16\cdot6}}{8} \\\\ x=\dfrac{4\pm i\sqrt{(4)^2\cdot6}}{8} \\\\ x=\dfrac{4\pm i(4)\sqrt{6}}{8} \\\\ x=\dfrac{4\pm 4i\sqrt{6}}{8} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel4(1)\pm \cancel4(1)i\sqrt{6}}{\cancel4(2)} \\\\ x=\dfrac{1\pm i\sqrt{6}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{1-i\sqrt{6}}{2} \\\\\text{OR}\\\\ x=\dfrac{1+i\sqrt{6}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{1-i\sqrt{6}}{2},\dfrac{1+i\sqrt{6}}{2} \right\} .$
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