Solve an equation involving $r$ in the denominator

Question

Sorry if the title is wrong, I am not sure what to call this. It's been a really long time since I've taken math, and I can't remember for the life of me how to solve this, nor can I figure out how to Google it. It should come out to $.20$ but I continue to get $.30$.

$100=\frac {70} {(1+r)} + \frac {60} {(1+r)^2} $

$ r = ? $

Answer 1

To reduce the amount of upfront work I'm just going to make the substitution $$t=\frac{1}{1+r}$$ This will give us the equation $$100=70t+60t^2 \text{ or, after reducing a bit } 10=7t+6t^2$$ Solving for $t$ we have $$\begin{array}{lll} 6t^2+7t-10&=&0\\ 6(6t^2+7t-10)&=&0\\ (6t)^2+7(6t)-60&=&0\\ (6t)^2+12(6t)-5(6t)-60&=&0\\ (6t)(6t+12)-5(6t+12)&=&0\\ (6t-5)(6t+12)&=&0\\ 6(6t-5)(t+2)&=&0\\ (6t-5)(t+2)&=&0\\ \end{array}$$ which gives $$t = \frac{5}{6}\text{ or } t=-2$$ or alternatively $$\frac{1}{t}=\frac{6}{5}\text{ or } \frac{1}{t}=-\frac{1}{2}$$ back substituting we have $$1+r = \frac{6}{5}\implies r=\frac{1}{5} \text{ or } 1+r=-\frac{1}{2}\implies r=-\frac{3}{2}$$

Answer 2

Let $u = 1 + r$. Then your equation takes the form $$100u^2 = 70u + 60.$$

This can be written in standard form as: $$100u^2 - 70u - 60 = 0$$ whose left-hand side can be factored as: $$10(5u - 6)(2u + 1) = 0.$$

Consequently, either $$5u - 6 = 0 \Longleftrightarrow u = 1 + r = \frac{6}{5} \Longleftrightarrow r = \frac{1}{5} = 0.2$$ or $$2u + 1 = 0 \Longleftrightarrow u = 1 + r = -\frac{1}{2} \Longleftrightarrow r = -\frac{3}{2}.$$

From your original post, I am guessing the root $r = -3/2$ is extraneous.

Answer 3

This outlines how to solve these types of equations using the Least Common Denominator (LCD). We are going to multiply everything by the LCD. To find the LCD, we look at the factors of each denominator and multiply the factors with the highest powers.

We have two denominators: $$(1 + r) \quad \mbox{ and } \quad (1+r)^2.$$ The only factor in the denominator is $(1+r)$, and the highest power of this is $(1+r)^2$. Thus, the LCD is $(1+r)^2$.

Now, multiply everything by $(1+r)^2$

$$100 \cdot (1+r)^2=\frac {70} {(1+r)} \cdot (1+r)^2+ \frac {60} {(1+r)^2}\cdot (1+r)^2.$$

A single $(1+r)$ will cancel out in the middle expression, and $(1+r)^2$ will cancel out in the last expression, leaving us with:

$$100 \cdot (1+r)^2=70 \cdot (1+r) + 60.$$

Since $(1+r)^2 = 1 + 2r + r^2$, we get:

$$100 \cdot (1 + 2r + r^2)=70 \cdot (1+r) + 60.$$

After this, distribute the $100$ and the $70$

$$ 100 + 200r + 100r^2 = 70 + 70r + 60.$$

Can you take it from here and solve this quadratic equation?