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: 37


$x=\left\{ \dfrac{2-i}{4},\dfrac{2+i}{4} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ (2x-1)(8x-4)=-1 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} 2x(8x)+2x(-4)-1(8x)-1(-4)=-1 \\\\ 16x^2-8x-8x+4=-1 \\\\ 16x^2-16x+4=-1 \\\\ 16x^2-16x+4+1=0 \\\\ 16x^2-16x+5=0 .\end{array} The quadratic equation above has $a= 16 , b= -16 , c= 5 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-16)\pm\sqrt{(-16)^2-4(16)(5)}}{2(16)} \\\\ x=\dfrac{16\pm\sqrt{256-320}}{32} \\\\ x=\dfrac{16\pm\sqrt{-64}}{32} .\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{16\pm\sqrt{-1}\cdot\sqrt{64}}{32} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{16\pm i\sqrt{64}}{32} .\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{16\pm i\sqrt{(8)^2}}{32} \\\\ x=\dfrac{16\pm 8i}{32} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel8(2)\pm \cancel8(1)i}{\cancel8(4)} \\\\ x=\dfrac{2\pm i}{4} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{2-i}{4} \\\\\text{OR}\\\\ x=\dfrac{2+i}{4} .\end{array} Hence, $ x=\left\{ \dfrac{2-i}{4},\dfrac{2+i}{4} \right\} .$
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