Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises: 12

Answer

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

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 2x^2=3-x ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} 2x^2+x-3=0 \\\\ (x-1)(2x+3) .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} x-1=0 \text{ OR } 2x+3=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 \\\\\text{ OR }\\\\ 2x+3=0 \\\\ 2x=-3 \\\\ x=-\dfrac{3}{2} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{3}{2},1 \right\} .$ $\bf{\text{Supplementary Solution:}}$ To factor the expression, $ 2x^2+x-3 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. In the expression above the value of $ac$ is $ 2(-3)=-6 $ and the value of $b$ is $ 1 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-6\}, \{2,-3\}, \{-1,6\}, \{-2,3\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -2,3 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2x^2-2x+3x-3 .\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} (2x^2-2x)+(3x-3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(x-1)+3(x-1) .\end{array} Factoring the $GCF= (x-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (x-1)(2x+3) .\end{array}
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