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: 44


$x=\left\{ -\dfrac{5}{3},1 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 3x^2+2x=5 ,$ using the completing the square method, ensure first that the coefficient of the squared variable is $1.$ Then add the quantity $\left( \dfrac{b}{2} \right)^2$ to both sides. Factor the resulting perfect square trinomial at the left side of the equal sign. Then take the square root of both sides. Finally, solve for the value of the variable. $\bf{\text{Solution Details:}}$ Dividing both sides by $ 3 $ so that the coefficient of the squared variable is $1$ results to \begin{array}{l}\require{cancel} x^2+\dfrac{2}{3}x=\dfrac{5}{3} .\end{array} Making the left side a perfect square trinomial by adding the quantity $ \left( \dfrac{b}{2} \right)^2 $ to both sides results to \begin{array}{l}\require{cancel} x^2+\dfrac{2}{3}x+\left( \dfrac{2/3}{2} \right)^2=\dfrac{5}{3}+\left( \dfrac{2/3}{2} \right)^2 \\\\ x^2+\dfrac{2}{3}x+\left( \dfrac{1}{3} \right)^2=\dfrac{5}{3}+\left( \dfrac{1}{3} \right)^2 \\\\ x^2+\dfrac{2}{3}x+\dfrac{1}{9}=\dfrac{5}{3}+\dfrac{1}{9} \\\\ \left( x+\dfrac{1}{3}\right)^2=\dfrac{15}{9}+\dfrac{1}{9} \\\\ \left( x+\dfrac{1}{3}\right)^2=\dfrac{16}{9} .\end{array} Taking the square root of both sides and then isolating the variable result to \begin{array}{l}\require{cancel} x+\dfrac{1}{3}=\pm\sqrt{\dfrac{16}{9}} \\\\ x+\dfrac{1}{3}=\pm\dfrac{4}{3} \\\\ x=-\dfrac{1}{3}\pm\dfrac{4}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=-\dfrac{1}{3}-\dfrac{4}{3} \\\\ x=-\dfrac{5}{3} ,\\\\\text{ OR }\\\\ x=-\dfrac{1}{3}+\dfrac{4}{3} \\\\ x=\dfrac{3}{3} \\\\ x=1 .\end{array} Hence, $ x=\left\{ -\dfrac{5}{3},1 \right\} .$
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