## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 37

#### Answer

$8x^{2}+6x+1=0$

#### Work Step by Step

$x=-\displaystyle \frac{1}{4}$ or $x=-\displaystyle \frac{1}{2}\qquad$....get $0$'s on one side. $x+\displaystyle \frac{1}{4}=0$ or $x+\displaystyle \frac{1}{2}=0\qquad$....Use the principle of zero products (multiplying). The solutions of this equation are $-\displaystyle \frac{1}{4}$ and $-\displaystyle \frac{1}{2}$. $(x+\displaystyle \frac{1}{4})(x+\frac{1}{2})=0\qquad$...use the FOIL method to multiply. $x^{2}+\displaystyle \frac{1}{2}x+\frac{1}{4}x+\frac{1}{8}=0\qquad$...combine like terms. $x^{2}+\displaystyle \frac{3}{4}x+\frac{1}{8}=0\qquad$...multiply both sides by $8$ to clear fractions. $8x^{2}+6x+1=0$

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