Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 697: 48

Answer

The solutions are $r=0$ or $r=-1$.

Work Step by Step

$\frac{2}{r}+\frac{1}{r^2}+\frac{r^2+r}{r^3}=\frac{1}{r}$ $\frac{2r^2+r+r^2+r}{r^3}=\frac{1}{r}$ $\frac{3r^2+2r}{r^3}=\frac{1}{r}$ $3r^3+2r^2=r^3$ $2r^3+2r^2=0$ $2r^2(r+1)$ $r=0$ or $r=-1$ Check: $\frac{2}{0}+\frac{1}{0^2}+\frac{0^2+r}{0^3}=\frac{1}{0}$ $0=0$ $\frac{2}{-1}+\frac{1}{(-1)^2}+\frac{(-1)^2+(-1)}{(-1)^3}=\frac{1}{(-1)}$ $-1=-1$ The solutions are $r=0$ or $r=-1$.

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