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 - Page 356: 31


$x=\left\{ -3,4 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 2x^2-12-4x=x^2-3x ,$ in $ax^2+bx+c=0$ form. Then, express the equation in factored form. Next, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ Using the properties of equality and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (2x^2-x^2)+(-4x+3x)-12=0 \\\\ x^2-x-12=0 .\end{array} To factor the trinomial expression above, find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ In the expression above, the value of $c$ is $ -12 $ and the value of $b$ is $ -1 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,-12 \}, \{ 2,-6 \}, \{ 3,-4 \}, \\ \{ -1,12 \}, \{ -2,6 \}, \{ -3,4 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 3,-4 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x+3)(x-4)=0 .\end{array} Equating each of the factors in the expression above to zero (Zero-Factor Theorem), then, \begin{array}{l}\require{cancel} x+3=0 \text{ OR } x-4=0 .\end{array} Solving each of the equations above results to \begin{array}{l}\require{cancel} x+3=0 \\\\ x=-3 \\\\\text{ OR }\\\\ x-4=0 \\\\ x=4 .\end{array} Hence, the solutions are $ x=\left\{ -3,4 \right\} .$
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