College Algebra (11th Edition)

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

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 17


$x=\left\{ -\dfrac{3}{4},1 \right\} $

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the given equation, $ -4x^2+x=-3 .$ Then equate each factor to zero and solve for the values of the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} -4x^2+x+3=0 \\\\ (-1)(-4x^2+x+3)=(0)(-1) \\\\ 4x^2-x-3=0 .\end{array} The two numbers whose product is $ac= 4(-3)=-12 $ and whose sum is $b= -1 $ are $\{ -4,3 \}$. Using these two numbers to decompose the middle term of the equation above results to \begin{array}{l}\require{cancel} 4x^2-4x+3x-3=0 \\\\ (4x^2-4x)+(3x-3)=0 \\\\ 4x(x-1)+3(x-1)=0 \\\\ (x-1)(4x+3)=0 .\end{array} Equating each factor to zero (or the Zero-Factor Property), the solutions to the given equation are \begin{array}{l}\require{cancel} x-1=0 \\\\ x=1 ,\\\\\text{OR}\\\\ 4x+3=0 \\\\ 4x=-3 \\\\ x=-\dfrac{3}{4} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{3}{4},1 \right\} .$
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