Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.2 - The Quadratic Formula - 8.2 Exercises: 21


$r=\left\{ -1-3\sqrt{2},-1+3\sqrt{2} \right\} $

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ (r-3)(r+5)=2 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} r(r)+r(5)-3(r)-3(5)=2 \\\\ r^2+5r-3r-15=2 \\\\ r^2+2r-15=2 .\end{array} Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to \begin{array}{l}\require{cancel} r^2+2r-15-2=0 \\\\ r^2+2r-17=0 .\end{array} The quadratic equation above has $a= 1 , b= 2 , c= -17 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} r=\dfrac{-2\pm\sqrt{2^2-4(1)(-17)}}{2(1)} \\\\ r=\dfrac{-2\pm\sqrt{4+68}}{2} \\\\ r=\dfrac{-2\pm\sqrt{72}}{2} .\end{array} Simplifying the radical by writing the radicand as an expression that contains a factor that is a perfect power of the index and then extracting the root of that factor result to \begin{array}{l}\require{cancel} r=\dfrac{-2\pm\sqrt{36\cdot2}}{2} \\\\ r=\dfrac{-2\pm\sqrt{(6)^2\cdot2}}{2} \\\\ r=\dfrac{-2\pm6\sqrt{2}}{2} .\end{array} Cancelling the common factors from all the terms results to \begin{array}{l}\require{cancel} r=\dfrac{\cancel2(-1)\pm\cancel2(3)\sqrt{2}}{\cancel2(1)} \\\\ r=-1\pm3\sqrt{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} r=-1-3\sqrt{2} \\\\\text{OR}\\\\ r=-1+3\sqrt{2} .\end{array} Hence, $ r=\left\{ -1-3\sqrt{2},-1+3\sqrt{2} \right\} .$
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