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 - Page 519: 20


$x=\left\{ \dfrac{-3-\sqrt{21}}{3},\dfrac{-3+\sqrt{21}}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ -3x(x+2)=-4 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -3x(x)-3x(2)=-4 \\\\ -3x^2-6x=-4 .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} -3x^2-6x+4=0 \\\\ -1(-3x^2-6x+4)=-1(0) \\\\ 3x^2+6x-4=0 .\end{array} The quadratic equation above has $a= 3 , b= 6 , c= -4 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-6\pm\sqrt{6^2-4(3)(-4)}}{2(3)} \\\\ x=\dfrac{-6\pm\sqrt{36+48}}{6} \\\\ x=\dfrac{-6\pm\sqrt{84}}{6} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} x=\dfrac{-6\pm\sqrt{4\cdot21}}{6} \\\\ x=\dfrac{-6\pm\sqrt{(2)^2\cdot21}}{6} \\\\ x=\dfrac{-6\pm2\sqrt{21}}{6} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel{2}(-3)\pm\cancel{2}(1)\sqrt{21}}{\cancel{2}(3)} \\\\ x=\dfrac{-3\pm\sqrt{21}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-\sqrt{21}}{3} \\\\\text{OR}\\\\ x=\dfrac{-3+\sqrt{21}}{3} .\end{array} Hence, $ x=\left\{ \dfrac{-3-\sqrt{21}}{3},\dfrac{-3+\sqrt{21}}{3} \right\} .$
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