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: 16


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

Work Step by Step

The two numbers whose product is $ac= 2(-15)=-30 $ and whose sum is $b= -1 $ are $\{ -6,5 \}$. Using these two numbers to decompose the middle term of the given equation, $ 2x^2-x-15=0 ,$ results to \begin{array}{l}\require{cancel} 2x^2-6x+5x-15=0 \\\\ (2x^2-6x)+(5x-15)=0 \\\\ 2x(x-3)+5(x-3)=0 \\\\ (x-3)(2x+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-3=0 \\\\ x=3 ,\\\\\text{OR}\\\\ 2x+5=0 \\\\ 2x=-5 \\\\ x=-\dfrac{5}{2} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{5}{2},3 \right\} .$
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