Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 46


$x=\left\{ \dfrac{1}{2},\dfrac{3}{5}\right\}$

Work Step by Step

Using the properties of equality, the given expression is equivalent to \begin{array}{l}\require{cancel} 3+10x^2=11x \\\\ 10x^2-11x+3=0 .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ equation }$ \begin{array}{l}\require{cancel} 10x^2-11x+3=0 \end{array} has $ac= 10(3)=30 $ and $b= -11 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -5,-6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 10x^2-5x-6x+3=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} (10x^2-5x)-(6x-3)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5x(2x-1)-3(2x-1)=0 .\end{array} Factoring the $GCF= (2x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2x-1)(5x-3)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solution is $ x=\left\{ \dfrac{1}{2},\dfrac{3}{5}\right\} .$
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