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: 64


$x=\left\{ \dfrac{2-\sqrt{10}}{3},\dfrac{2+\sqrt{10}}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ (3x+2)(x-1)=3x ,$ use the FOIL Method and express the equation 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} 3x(x)+3x(-1)+2(x)+2(-1)=3x \\\\ 3x^2-3x+2x-2=3x \\\\ 3x^2+(-3x+2x-3x)-2=0 \\\\ 3x^2-4x-2=0 .\end{array} In the equation above, $a= 3 ,$ $b= -4 ,$ and $c= -2 .$ 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(3)(-2)}}{2(3)} \\\\ x=\dfrac{4\pm\sqrt{16+24}}{6} \\\\ x=\dfrac{4\pm\sqrt{40}}{6} .\end{array} Extracting the perfect square factor of the radicand, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{4\pm\sqrt{4\cdot10}}{6} \\\\ x=\dfrac{4\pm\sqrt{(2)^2\cdot10}}{6} \\\\ x=\dfrac{4\pm2\sqrt{10}}{6} \\\\ x=\dfrac{2(2\pm\sqrt{10})}{6} \\\\ x=\dfrac{\cancel2(2\pm\sqrt{10})}{\cancel2(3)} \\\\ x=\dfrac{2\pm\sqrt{10}}{3} .\end{array} The solutions are $ x=\left\{ \dfrac{2-\sqrt{10}}{3},\dfrac{2+\sqrt{10}}{3} \right\} .$
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