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: 18


$x=\left\{ -\dfrac{5}{6},2 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the given equation, $ -6x^2+7x=-10 .$ 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} -6x^2+7x+10=0 \\\\ -1(-6x^2+7x+10)=(0)(-1) \\\\ 6x^2-7x-10=0 .\end{array} The two numbers whose product is $ac= 6(-10)=-60 $ and whose sum is $b= -7 $ are $\{ -12,5 \}$. Using these two numbers to decompose the middle term of the equation above results to \begin{array}{l}\require{cancel} 6x^2-12x+5x-10=0 \\\\ (6x^2-12x)+(5x-10)=0 \\\\ 6x(x-2)+5(x-2)=0 \\\\ (x-2)(6x+5)=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-2=0 \\\\ x=2 ,\\\\\text{OR}\\\\ 6x+5=0 \\\\ 6x=-5 \\\\ x=-\dfrac{5}{6} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{5}{6},2 \right\} .$
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