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: 64

Answer

$x=\left\{ \dfrac{-13-\sqrt{181}}{2},\dfrac{-13+\sqrt{181}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^2+13x-3=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+13x=3 .\end{array} In the equation above, $b= 13 .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{13}{2} \right)^2 \\\\= \dfrac{169}{4} .\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+13x+\dfrac{169}{4}=3+\dfrac{169}{4} \\\\ x^2+13x+\dfrac{169}{4}=\dfrac{12}{4}+\dfrac{169}{4} \\\\ x^2+13x+\dfrac{169}{4}=\dfrac{181}{4} .\end{array} With the left side now a perfect square trinomial, the equation above is equivalent to \begin{array}{l}\require{cancel} \left( x+\dfrac{13}{2} \right)^2=\dfrac{181}{4} .\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+\dfrac{13}{2}=\pm\sqrt{\dfrac{181}{4}} \\\\ x+\dfrac{13}{2}=\pm\sqrt{\dfrac{1}{4}\cdot181} \\\\ x+\dfrac{13}{2}=\pm\dfrac{1}{2}\sqrt{181} \\\\ x+\dfrac{13}{2}=\pm\dfrac{\sqrt{181}}{2} \\\\ x=-\dfrac{13}{2}\pm\dfrac{\sqrt{181}}{2} \\\\ x=\dfrac{-13\pm\sqrt{181}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-13-\sqrt{181}}{2} \\\\\text{OR}\\\\ x=\dfrac{-13+\sqrt{181}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{-13-\sqrt{181}}{2},\dfrac{-13+\sqrt{181}}{2} \right\} .$
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