College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 57


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ -4x^2=-12x+11 ,$ express the equation 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+12x-11=0 \\\\ \dfrac{-4x^2+12x-11}{-4}=\dfrac{0}{-4} \\\\ x^2-3x+\dfrac{11}{4}=0 .\end{array} In the equation above, $a= 1 ,$ $b= -3 ,$ and $c= \dfrac{11}{4} .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)\left(\dfrac{11}{4}\right)}}{2(1)} \\\\ x=\dfrac{3\pm\sqrt{9-11}}{2} \\\\ x=\dfrac{3\pm\sqrt{-2}}{2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and that $i=\sqrt{-1},$ the expression above is equivalent to\begin{array}{l}\require{cancel} x=\dfrac{3\pm\sqrt{-1}\cdot\sqrt{2}}{2} \\\\= x=\dfrac{3\pm i\sqrt{2}}{2} .\end{array} The solutions are $ x=\left\{ \dfrac{3-i\sqrt{2}}{2},\dfrac{3+i\sqrt{2}}{2} \right\} .$
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