Solve for r

S

Sirch

New member
Joined
Jun 12, 2009
Messages
2
I have a question that says

Solve for r
SQRT((r+f)/(r-f))=a/d

Now, when I type this into any number of online solvers the answer is
r = (-f(a^2+d^2))/(a^2-d^2)

I've got as far as
(r+f)/(r-f)=(a/d)^2
but that's about it.

I want to know how to do it, because I have a few more questions like this.
Thanks for your help.
 
Sirch said:
I have a question that says

Solve for r
SQRT((r+f)/(r-f))=a/d

Now, when I type this into any number of online solvers the answer is
r = (-f(a^2+d^2))/(a^2-d^2)

I've got as far as
(r+f)/(r-f)=(a/d)^2
but that's about it.

I want to know how to do it, because I have a few more questions like this.
Thanks for your help.
Click to expand...

Ok, so far, so good:

(r + f) / (r - f) = (a/d)[sup:2ve301rg]2[/sup:2ve301rg]

or,

(r + f) / (r - f) = a[sup:2ve301rg]2[/sup:2ve301rg] / d[sup:2ve301rg]2[/sup:2ve301rg]

Cross-multiply:

d[sup:2ve301rg]2[/sup:2ve301rg]*(r + f) = a[sup:2ve301rg]2[/sup:2ve301rg]*(r - f)
d[sup:2ve301rg]2[/sup:2ve301rg]r + d[sup:2ve301rg]2[/sup:2ve301rg]f = a[sup:2ve301rg]2[/sup:2ve301rg]r - a[sup:2ve301rg]2[/sup:2ve301rg]f

Since you are trying to solve for "r" I would get all terms containing r on one side of the equation, and everything else on the other side.

d[sup:2ve301rg]2[/sup:2ve301rg] r - a[sup:2ve301rg]2[/sup:2ve301rg]r = -d[sup:2ve301rg]2[/sup:2ve301rg]f - a[sup:2ve301rg]2[/sup:2ve301rg]f

Ok...now, factor r out of both terms on the left side and f out of both terms on the right:

r(d[sup:2ve301rg]2[/sup:2ve301rg] - a[sup:2ve301rg]2[/sup:2ve301rg]) = f(-d[sup:2ve301rg]2[/sup:2ve301rg] - a[sup:2ve301rg]2[/sup:2ve301rg])

Divide both sides by( d[sup:2ve301rg]2[/sup:2ve301rg] - a[sup:2ve301rg]2[/sup:2ve301rg])

Ok...I've gotten you REAL close...you can finish it, I hope.
 
Hello, Sirch!

\(\displaystyle \text{Solve for }r\!:\quad \sqrt{\frac{r+f}{r-f}} \:=\:\frac{a}{d}\)
Click to expand...

\(\displaystyle \text{Square both sides: }\;\frac{r+f}{r-f} \;=\;\frac{a^2}{d^2} \quad\Rightarrow\quad d^2(r+f) \;=\;a^2(r-f)\)

. . \(\displaystyle d^2r + d^2f \;=\;a^2r - a^2f \quad\Rightarrow\quad -a^2r + d^2r \;=\;-a^2f - d^2f\)

\(\displaystyle \text{Factor: }\;\;-r(a^2-d^2) \;=\;-f(a^2+d^2) \quad\Rightarrow\quad r \;=\;\frac{-f(a^2+d^2)}{-(a^2-d^2)}\)


\(\displaystyle \text{Therefore: }\;\;r \;=\;\frac{f(a^2+d^2)}{a^2-d^2}\)


Edit: Too slow again . . . *sigh*
.
 
\(\displaystyle \sqrt{\frac{r+f}{r-f}}=\frac{a}{d}\)

Square both sides, as you done:

\(\displaystyle \frac{r+f}{r-f}=\frac{a^{2}}{d^{2}}\)

Long divide the left side and rewrite:

\(\displaystyle \frac{2f}{r-f}=\frac{a^{2}}{d^{2}}-1\)

Reciprocal:

\(\displaystyle \frac{r-f}{2f}=\frac{d^{2}}{a^{2}-d^{2}}\)

\(\displaystyle r-f=\frac{2fd^{2}}{a^{2}-d^{2}}\)

\(\displaystyle r=\frac{2fd^{2}}{a^{2}-d^{2}}+f=\frac{f(a^{2}+d^{2})}{a^{2}-d^{2}}\)
 
You must log in or register to reply here.
Top