College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 164: 66


$x=\left\{ -\dfrac{1}{2}, 3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2-\dfrac{5}{x}=\dfrac{3}{x^2} ,$ multiply both sides by the $LCD.$ Then express the resulting equation in the form $ax^2+bx+c=0.$ Use concepts of solving quadratic equations to find the values of $x.$ Finally, do checking and ensure that any denominator does not become $0.$ $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $1,x,$ and $x^2$ is $x^2$ since it is the lowest expression which can be divided exactly by all the given denominators. Multiplying both sides by the $LCD= x^2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} x^2\left( 2-\dfrac{5}{x} \right)=\left( \dfrac{3}{x^2} \right)x^2 \\\\ x^2(2)-x(5)=1(3) \\\\ 2x^2-5x=3 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2x^2-5x-3=0 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is \begin{array}{l}\require{cancel} (2x+1)(x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property) results to \begin{array}{l}\require{cancel} 2x+1=0 \\\\\text{OR}\\\\ x-3=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+1=0 \\\\ 2x=-1 \\\\ x=-\dfrac{1}{2} \\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Upon checking, the solutions do not produce an expression with a $0$ in the denominator. Hence, the solutions are $ x=\left\{ -\dfrac{1}{2}, 3 \right\} .$
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