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: 75

Answer

$ 10\%$

Work Step by Step

$ A=P(1+r)^{t}\qquad$..Substitute $2420$ for $A,\ 2000$ for $P$ and $2$ for $t$. $ 2420=2000(1+r)^{2}\qquad$..divide both sides by $2000$. $\displaystyle \frac{2420}{2000}=(1+r)^{2}\qquad$..simplify. $\displaystyle \frac{121}{100}=(1+r)^{2}$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $ 1+r=\pm\sqrt{\frac{121}{100}}\qquad$..add $-1$ to both sides. $ 1-1+r=\displaystyle \pm\frac{\sqrt{121}}{\sqrt{100}}-1\qquad$..simplify. $ r=-1\displaystyle \pm\frac{11}{10}\qquad$.. The interest rate cannot be negative, we eliminate the negative solution. $ r=-1+\displaystyle \frac{11}{10}=0.1=10\%$
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