# Get in terms of d

1. Feb 24, 2006

### UrbanXrisis

I dont know why I cant do this, all I need to do is to get the variable d on one side and everything else on the other side:

constants: the f's and s
variables: l and d

$$\frac{1}{\frac{1}{\frac{1}{l}-\frac{1}{f_2}}+d}-\frac{1}{d+l-s}=\frac{1}{f_1}$$

$$\frac{f_2-l}{df_2-dl+lf_2}-\frac{1}{d+l-s}=\frac{1}{f_1}$$

$$\frac{d_f2+f_2l-f_2s-f_2s-dl-l^2+ls-df_2+dl-lf_2}{d^2f-d^2l+dlf_2+df_2l-dl^2+f_2l^2-sdf_2+dls-slf_2}=\frac{1}{f_1}$$

when I try to solve for d, i get d^2(...)+d(....)=constants
I cant solve this because both sides will have a d in it, is there something I'm missing?

Last edited: Feb 24, 2006
2. Feb 24, 2006

### Hurkyl

Staff Emeritus
What kind of equation is d^2(...)+d(....)=constants?

I bet you know how to solve (...) x^2 + (...) x + (...) = 0.

3. Feb 24, 2006

### UrbanXrisis

yes but the (...) has the variable l in it (l is not a constant), so I cant solve for d. I need to solve for d in terms of l

4. Feb 24, 2006

### Hurkyl

Staff Emeritus
So? How does that change anything?

5. Feb 24, 2006

### quasar987

What matters is that l is not a function of d.

What I'm saying is, if x and t are two independant variables, and we have

$$f(t)x^2+g(t)x+h(t)=0$$

then the solutions for x are a function of t:

$$x_{1,2}(t) = \frac{-g(t) \pm \sqrt{g^2(t) - 4f(t)h(t)}}{2f(t)}$$