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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 707: 65


The solutions are $x=-\displaystyle \frac{4}{3}$ and $x=-\displaystyle \frac{2}{3}$.

Work Step by Step

$ 9x^{2}+18x=-8\qquad$..divide both sides with $9$ $ x^{2}+2x=-\displaystyle \frac{8}{9}\qquad$..add $1$ to both sides to complete the square ($\displaystyle \frac{1}{2}(2)=1$, and $(1)^{2}=1$.) $ x^{2}+2x+1=-\displaystyle \frac{8}{9}+1\qquad$...simplify by applying the Perfect square formula ($(x+a)^{2}=x^{2}+2ax+a^{2}$) and adding like terms. $(x+1)^{2}=\displaystyle \frac{1}{9}$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $ x+1=\pm\sqrt{\frac{1}{9}}\qquad$...add $-1$ to each side. $ x+1-1=\pm\sqrt{\frac{1}{9}}-1\qquad$...simplify. $ x=-1\pm\sqrt{\frac{1}{9}}\qquad$...apply the quotient rule for radicals: $\displaystyle \sqrt{\frac{m}{n}}=\frac{\sqrt{m}}{\sqrt{n}}$ and simplify $x=-1+\displaystyle \frac{1}{3}$ or $x=-1-\displaystyle \frac{1}{3}$ $x=-\displaystyle \frac{2}{3}$ or $x=-\displaystyle \frac{4}{3}$
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