College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - 1.6 Exercises: 25

Answer

$x=\left\{ -2,\dfrac{5}{4} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given rational equation, $ \dfrac{2x}{x-2}=5+\dfrac{4x^2}{x-2} ,$ multiply both sides by the $LCD.$ Then use concepts of quadratic equations to solve for the variable. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators is the collection of all not common factors with the highest exponent and the common factors with the highest exponent. Hence the $LCD$ of $\left\{ (x-2),1,(x-2) \right\}$ is $ (x-2) .$ Multiplying both sides of the given equation by the $LCD= (x-2) $ results to \begin{array}{l}\require{cancel} (x-2)\left(\dfrac{2x}{x-2}\right)=\left(5+\dfrac{4x^2}{x-2}\right)(x-2) .\end{array} Using the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (x-2)\cdot\dfrac{2x}{x-2}=5\cdot(x-2)+\dfrac{4x^2}{x-2}\cdot(x-2) \\\\ (\cancel{x-2})\cdot\dfrac{2x}{\cancel{x-2}}=5\cdot(x-2)+\dfrac{4x^2}{\cancel{x-2}}\cdot(\cancel{x-2}) \\\\ 1(2x)=5(x-2)+(4x^2)(1) \\\\ 2x=5x-10+4x^2 \\\\ -4x^2+(2x-5x)+10=0 \\\\ -4x^2-3x+10=0 \\\\ -1(-4x^2-3x+10)=(0)(-1) \\\\ 4x^2+3x-10=0 .\end{array} In the trinomial expression above the value of $ac$ is $ 4(-10)=-40 $ and the value of $b$ is $ 3 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -5,8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 4x^2-5x+8x-10=0 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (4x^2-5x)+(8x-10)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x(4x-5)+2(4x-5)=0 .\end{array} Factoring the $GCF= (4x-5) $ of the entire expression above results to \begin{array}{l}\require{cancel} (4x-5)(x+2)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 4x-5=0 \\\\\text{OR}\\\\ x+2=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 4x-5=0 \\\\ 4x=5 \\\\ x=\dfrac{5}{4} \\\\\text{OR}\\\\ x+2=0 \\\\ x=-2 .\end{array} Upon checking, none of the solutions above will make the denominator of the original equation equal to zero. Hence, $ x=\left\{ -2,\dfrac{5}{4} \right\} .$
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