College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 164: 68


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{13}{x^2+10}=\dfrac{2}{x} ,$ use cross-multiplication. Then use the concepts of solving quadratic equations to find the values of $x.$ Finally, do checking and ensure that any denominator does not become $0.$ $\bf{\text{Solution Details:}}$ 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} 13(x)=(x^2+10)(2) \\\\ 13x=2x^2+20 .\end{array} In the form $ax^2+bx+c=0,$ the equation above is equivalent to \begin{array}{l}\require{cancel} -2x^2+13x-20=0 \\\\ -1(-2x^2+13x-20)=-1(0) \\\\ 2x^2-13x+20=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-5)(x-4)=0 .\end{array} Equating each factor to zero (Zero Product Property) results to \begin{array}{l}\require{cancel} 2x-5=0 \\\\\text{OR}\\\\ x-4=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x-5=0 \\\\ 2x=5 \\\\ x=\dfrac{5}{2} \\\\\text{OR}\\\\ x-4=0 \\\\ x=4 .\end{array} Upon checking, the solutions do not produce an expression with a $0$ in the denominator. Hence, the solutions are $ x=\left\{ \dfrac{5}{2}, 4 \right\} .$
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