## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises - Page 512: 63

#### Answer

$x=\left\{ \dfrac{-7-\sqrt{53}}{2},\dfrac{-7+\sqrt{53}}{2} \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $x^2+7x-1=0 ,$ 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+7x=1 .\end{array} In the equation above, $b= 7 .$ The expression $\left( \dfrac{b}{2} \right)^2,$ evaluates to \begin{array}{l}\require{cancel} \left( \dfrac{7}{2} \right)^2 \\\\= \dfrac{49}{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+7x+\dfrac{49}{4}=1+\dfrac{49}{4} \\\\ x^2+7x+\dfrac{49}{4}=\dfrac{4}{4}+\dfrac{49}{4} \\\\ x^2+7x+\dfrac{49}{4}=\dfrac{53}{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{7}{2} \right)^2=\dfrac{53}{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{7}{2}=\pm\sqrt{\dfrac{53}{4}} \\\\ x+\dfrac{7}{2}=\pm\sqrt{\dfrac{1}{4}\cdot53} \\\\ x+\dfrac{7}{2}=\pm\dfrac{1}{2}\sqrt{53} \\\\ x+\dfrac{7}{2}=\pm\dfrac{\sqrt{53}}{2} \\\\ x=-\dfrac{7}{2}\pm\dfrac{\sqrt{53}}{2} \\\\ x=\dfrac{-7\pm\sqrt{53}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-7-\sqrt{53}}{2} \\\\\text{OR}\\\\ x=\dfrac{-7+\sqrt{53}}{2} .\end{array} Hence, $x=\left\{ \dfrac{-7-\sqrt{53}}{2},\dfrac{-7+\sqrt{53}}{2} \right\} .$

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