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: 62

Answer

$z=\left\{ -3,\dfrac{13}{4} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 4z^2-z=39 ,$ 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{4z^2-z}{4}=\dfrac{39}{4} \\\\ z^2-\dfrac{1}{4}z=\dfrac{39}{4} .\end{array} In the equation above, $b= -\dfrac{1}{4} .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{-\dfrac{1}{4}}{2} \right)^2 \\\\= \left( -\dfrac{1}{4}\div2 \right)^2 \\\\= \left( -\dfrac{1}{4}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( -\dfrac{1}{8} \right)^2 \\\\= \dfrac{1}{64} .\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} z^2-\dfrac{1}{4}z+\dfrac{1}{64}=\dfrac{39}{4}+\dfrac{1}{64} \\\\ z^2-\dfrac{1}{4}z+\dfrac{1}{64}=\dfrac{624}{64}+\dfrac{1}{64} \\\\ z^2-\dfrac{1}{4}z+\dfrac{1}{64}=\dfrac{625}{64} .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( z-\dfrac{1}{8} \right)^2=\dfrac{625}{64} .\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} z-\dfrac{1}{8}=\pm\sqrt{\dfrac{625}{64}} \\\\ z-\dfrac{1}{8}=\pm\dfrac{25}{8} \\\\ z=\dfrac{1}{8}\pm\dfrac{25}{8} .\end{array} The solutions are \begin{array}{l}\require{cancel} z=\dfrac{1}{8}-\dfrac{25}{8} \\\\ z=-\dfrac{24}{8} \\\\ z=-3 \\\\\text{OR}\\\\ z=\dfrac{1}{8}+\dfrac{25}{8} \\\\ z=\dfrac{26}{8} \\\\ z=\dfrac{13}{4} .\end{array} Hence, $ z=\left\{ -3,\dfrac{13}{4} \right\} .$
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