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: 59

Answer

$x=\left\{ -5-\sqrt{7},-5+\sqrt{7} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^2+10x+18=0 ,$ 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} x^2+10x=-18 .\end{array} In the equation above, $b= 10 .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{10}{2} \right)^2 \\\\= \left( 5 \right)^2 \\\\= 25 .\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} x^2+10x+25=-18+25 \\\\ x^2+10x+25=7 .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} (x+5)^2=7 .\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} x+5=\pm\sqrt{7} \\\\ x=-5\pm\sqrt{7} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=-5-\sqrt{7} \\\\\text{OR}\\\\ x=-5+\sqrt{7} .\end{array} Hence, $ x=\left\{ -5-\sqrt{7},-5+\sqrt{7} \right\} .$
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