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: 15

Answer

$x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ 2-2x=3x^2 ,$ express in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ 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^2-2x+2=0 \\\\ \dfrac{-3x^2-2x+2}{-1}=\dfrac{0}{-1} \\\\ 3x^2+2x-2=0 .\end{array} The quadratic equation above has $a= 3 , b= 2 , c= -2 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-2\pm\sqrt{2^2-4(3)(-2)}}{2(3)} \\\\ x=\dfrac{-2\pm\sqrt{4+24}}{6} \\\\ x=\dfrac{-2\pm\sqrt{28}}{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{-2\pm\sqrt{4\cdot7}}{6} \\\\ x=\dfrac{-2\pm\sqrt{(2)^2\cdot7}}{6} \\\\ x=\dfrac{-2\pm2\sqrt{7}}{6} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel2(-1)\pm\cancel2(1)\sqrt{7}}{\cancel2(3)} \\\\ x=\dfrac{-1\pm\sqrt{7}}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-1-\sqrt{7}}{3} \\\\\text{OR}\\\\ x=\dfrac{-1+\sqrt{7}}{3} .\end{array} Hence, $ x=\left\{ \dfrac{-1-\sqrt{7}}{3},\dfrac{-1+\sqrt{7}}{3} \right\} .$
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