Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises: 61

Answer

$w=\left\{ -\dfrac{8}{3},3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 3w^2-w=24 ,$ use first the properties of equality to express the equation in the form $x^2\pm bx=c.$ Once in this form, complete the square by adding $\left( \dfrac{b}{2} \right)^2$ to both sides of the equal sign. Then express the left side as a square of a binomial while simplify the right side. Then take the square root of both sides (Square Root Property) and use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, in the form $x^2+bx=c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{3w^2-w}{3}=\dfrac{24}{3} \\\\ w^2-\dfrac{1}{3}w=8 .\end{array} In the equation above, $b= -\dfrac{1}{3} .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{-\dfrac{1}{3}}{2} \right)^2 \\\\= \left( -\dfrac{1}{3}\div2 \right)^2 \\\\= \left( -\dfrac{1}{3}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( -\dfrac{1}{6} \right)^2 \\\\= \dfrac{1}{36} .\end{array} Adding the value of $\left( \dfrac{b}{2} \right)^2,$ to both sides of the equation above results to \begin{array}{l}\require{cancel} w^2-\dfrac{1}{3}w+\dfrac{1}{36}=8+\dfrac{1}{36} \\\\ w^2-\dfrac{1}{3}w+\dfrac{1}{36}=\dfrac{288}{36}+\dfrac{1}{36} \\\\ w^2-\dfrac{1}{3}w+\dfrac{1}{36}=\dfrac{289}{36} .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( w-\dfrac{1}{6} \right)^2=\dfrac{289}{36} .\end{array} Taking the square root of both sides (Square Root Property), simplifying the radical and then isolating the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} w-\dfrac{1}{6}=\pm\sqrt{\dfrac{289}{36}} \\\\ w-\dfrac{1}{6}=\pm\dfrac{17}{6} \\\\ w=\dfrac{1}{6}\pm\dfrac{17}{6} .\end{array} The solutions are \begin{array}{l}\require{cancel} w=\dfrac{1}{6}-\dfrac{17}{6} \\\\ w=-\dfrac{16}{6} \\\\ w=-\dfrac{8}{3} \\\\\text{OR}\\\\ w=\dfrac{1}{6}+\dfrac{17}{6} \\\\ w=\dfrac{18}{6} \\\\ w=3 .\end{array} Hence, $ w=\left\{ -\dfrac{8}{3},3 \right\} .$
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