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: 84


$x=\left\{ \dfrac{-5-i\sqrt{55}}{8},\dfrac{-5+i\sqrt{55}}{8} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $ 4x^2+5x+5=0 ,$ use the properties of equality to express the given equation in the form $x^2+bx=c.$ Then complete the square by adding $\left(\dfrac{b}{2} \right)^2$ to both sides. Factor the left side then take the square root (Square Root Property) of both sides. Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical and 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{4x^2+5x+5}{4}=\dfrac{0}{4} \\\\ x^2+\dfrac{5}{4}x+\dfrac{5}{4}=0 \\\\ x^2+\dfrac{5}{4}x=-\dfrac{5}{4} .\end{array} In the equation above, $b= \dfrac{5}{4} .$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then \begin{array}{l}\require{cancel} \left( \dfrac{\dfrac{5}{4}}{2} \right)^2 \\\\= \left( \dfrac{5}{4}\div2 \right)^2 \\\\= \left( \dfrac{5}{4}\cdot\dfrac{1}{2} \right)^2 \\\\= \left( \dfrac{5}{8} \right)^2 \\\\= \dfrac{25}{64} .\end{array} Adding $\left(\dfrac{b}{2} \right)^2$ to both sides of the equation above to complete the square, the equation becomes \begin{array}{l}\require{cancel} x^2+\dfrac{5}{4}x+\dfrac{25}{64}=-\dfrac{5}{4}+\dfrac{25}{64} \\\\ \left( x+\dfrac{5}{8} \right)^2=-\dfrac{80}{64}+\dfrac{25}{64} \\\\ \left( x+\dfrac{5}{8} \right)^2=-\dfrac{55}{64} .\end{array} Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} x+\dfrac{5}{8}=\pm\sqrt{-\dfrac{55}{64}} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x+\dfrac{5}{8}=\pm\sqrt{-1}\cdot\sqrt{\dfrac{55}{64}} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x+\dfrac{5}{8}=\pm i\sqrt{\dfrac{55}{64}} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} x+\dfrac{5}{8}=\pm i\sqrt{\dfrac{1}{64}\cdot55} \\\\ x+\dfrac{5}{8}=\pm i\sqrt{\left(\dfrac{1}{8}\right)^2\cdot55} \\\\ x+\dfrac{5}{8}=\pm i\left(\dfrac{1}{8}\right)\sqrt{55} \\\\ x+\dfrac{5}{8}=\pm\dfrac{i\sqrt{55}}{8} .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x=-\dfrac{5}{8}\pm\dfrac{i\sqrt{55}}{8} \\\\ x=\dfrac{-5\pm i\sqrt{55}}{8} .\end{array} Hence, $ x=\left\{ \dfrac{-5-i\sqrt{55}}{8},\dfrac{-5+i\sqrt{55}}{8} \right\} .$
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