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: 24



Work Step by Step

The two numbers whose product is $ac= 36(25)=900 $ and whose sum is $b= 60 $ are $\{ 30,30 \}$. Using these two numbers to decompose the middle term of the given equation, $ 36x^2+60x+25=0 ,$ results to \begin{array}{l}\require{cancel} 36x^2+30x+30x+25=0 \\\\ (36x^2+30x)+(30x+25)=0 \\\\ 6x(6x+6)+5(6x+5)=0 \\\\ (6x+6)(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} 6x+5=0 \\\\ 6x=-5 \\\\ x=-\dfrac{5}{6} ,\\\\\text{OR}\\\\ 6x+5=0 \\\\ 6x=-5 \\\\ x=-\dfrac{5}{6} .\end{array} Hence, the solution is $ x=-\dfrac{5}{6} .$
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