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 - Page 512: 71

Answer

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

Work Step by Step

In the form $x^2+bx=c,$ to complete the square of the expression at the left side, add $\left(\dfrac{b}{2}\right)^2$ to both sides of the equation. Thus, the given equation, $ z^2-\dfrac{4}{3}z=-\dfrac{1}{9} ,$ is equivalent to \begin{align*} z^2-\dfrac{4}{3}z+\left(\dfrac{-4/3}{2}\right)^2&=-\dfrac{1}{9}+\left(\dfrac{-4/3}{2}\right)^2 \\\\ z^2-\dfrac{4}{3}z+\left(\dfrac{-2}{3}\right)^2&=-\dfrac{1}{9}+\left(\dfrac{-2}{3}\right)^2 \\\\ z^2-\dfrac{4}{3}z+\dfrac{4}{9}&=-\dfrac{1}{9}+\dfrac{4}{9} \\\\ z^2-\dfrac{4}{3}z+\dfrac{4}{9}&=\dfrac{3}{9} .\end{align*} Using $a^2\pm2ab+b^2=(a\pm b)^2$ or the factoring of perfect square trinomials, the equation above is equivalent to \begin{align*} \left(z-\dfrac{2}{3}\right)^2&=\dfrac{3}{9} .\end{align*} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{align*} z-\dfrac{2}{3}&=\pm\sqrt{\dfrac{3}{9}} \\\\ z-\dfrac{2}{3}&=\pm\sqrt{\dfrac{1}{9}\cdot3} \\\\ z-\dfrac{2}{3}&=\pm\sqrt{\left(\dfrac{1}{3}\right)^2\cdot3} \\\\ z-\dfrac{2}{3}&=\pm\dfrac{1}{3}\sqrt{3} \\\\ z-\dfrac{2}{3}&=\pm\dfrac{\sqrt{3}}{3} .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*} z&=\dfrac{2}{3}\pm\dfrac{\sqrt{3}}{3} \\\\ z&=\dfrac{2\pm\sqrt{3}}{3} .\end{align*} Hence, the solution set of the equation $ z^2-\dfrac{4}{3}z=-\dfrac{1}{9} $ is $\left\{\dfrac{2-\sqrt{3}}{3},\dfrac{2+\sqrt{3}}{3}\right\}$.
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