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: 86

Answer

$x=\left\{ \dfrac{-5-i\sqrt{15}}{2},\dfrac{-5+i\sqrt{15}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $ -x^2-5x-10=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{-x^2-5x-10}{-1}=\dfrac{0}{-1} \\\\ x^2+5x+10=0 \\\\ x^2+5x=-10 .\end{array} In the equation above, $b= 5 .$ Substituting $b$ in the expression $\left( \dfrac{b}{2} \right)^2,$ then \begin{array}{l}\require{cancel} \left( \dfrac{5}{2} \right)^2 \\\\= \dfrac{25}{4} .\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+5x+\dfrac{25}{4}=-10+\dfrac{25}{4} \\\\ \left( x+\dfrac{5}{2} \right)^2=-\dfrac{40}{4}+\dfrac{25}{4} \\\\ \left( x+\dfrac{5}{2} \right)^2=-\dfrac{15}{4} .\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}{2}=\pm\sqrt{-\dfrac{15}{4}} .\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}{2}=\pm\sqrt{-1}\cdot\sqrt{\dfrac{15}{4}} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x+\dfrac{5}{2}=\pm i\sqrt{\dfrac{15}{4}} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect factor of the index and extracting its root, the equation above is equivalent to \begin{array}{l}\require{cancel} x+\dfrac{5}{2}=\pm i\sqrt{\dfrac{1}{4}\cdot15} \\\\ x+\dfrac{5}{2}=\pm i\sqrt{\left(\dfrac{1}{2}\right)^2\cdot15} \\\\ x+\dfrac{5}{2}=\pm i\left(\dfrac{1}{2}\right)\sqrt{15} \\\\ x+\dfrac{5}{2}=\pm \dfrac{i\sqrt{15}}{2} .\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}{2}\pm \dfrac{i\sqrt{15}}{2} \\\\ x=\dfrac{-5\pm i\sqrt{15}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{-5-i\sqrt{15}}{2},\dfrac{-5+i\sqrt{15}}{2} \right\} .$
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