College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 163: 46


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (x+4)(x+2)=2x ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula to solve for $x.$ $\bf{\text{Solution Details:}}$ 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} x(x)+x(2)+4(x)+4(2)=2x \\\\ x^2+2x+4x+8=2x .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x^2+(2x+4x-2x)+8=0 \\\\ x^2+4x+8=0 .\end{array} In the equation above, $a= 1 ,$ $b= 4 ,$ and $c= 8 .$ 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{-4\pm\sqrt{4^2-4(1)(8)}}{2(1)} \\\\ x=\dfrac{-4\pm\sqrt{16-32}}{2} \\\\ x=\dfrac{-4\pm\sqrt{-16}}{2} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy}$ and $i=\sqrt{-1},$ the equation above is equivalent to\begin{array}{l}\require{cancel} x=\dfrac{-4\pm\sqrt{-1}\cdot\sqrt{16}}{2} \\\\ x=\dfrac{-4\pm i\cdot\sqrt{(4)^2}}{2} \\\\ x=\dfrac{-4\pm 4i}{2} \\\\ x=\dfrac{\cancel2^{-2}\pm \cancel2^2i}{\cancel2^1} \\\\ x=-2\pm 2i .\end{array} The solutions are $ x=\left\{ -2-2i,-2+ 2i \right\} .$
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