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 - Page 519: 26


$x=\left\{ \dfrac{-3-\sqrt{33}}{2},\dfrac{-3+\sqrt{33}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $ x=\dfrac{2(x+3)}{x+5} ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{2(x)+2(3)}{x+5} \\\\ x=\dfrac{2x+6}{x+5} .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} x(x+5)=1(2x+6) \\\\ x^2+5x=2x+6 .\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} x^2+5x-2x-6=0 \\\\ x^2+3x-6=0 .\end{array} The quadratic equation above has $a= 1 , b= 3 , c= -6 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-3\pm\sqrt{3^2-4(1)(-6)}}{2(1)} \\\\ x=\dfrac{-3\pm\sqrt{9+24}}{2} \\\\ x=\dfrac{-3\pm\sqrt{33}}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-\sqrt{33}}{2} \\\\\text{OR}\\\\ x=\dfrac{-3+\sqrt{33}}{2} .\end{array} Hence, $ x=\left\{ \dfrac{-3-\sqrt{33}}{2},\dfrac{-3+\sqrt{33}}{2} \right\} .$
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