# Thread: Solve for R

1. ## Solve for R

p = [E^(2) R]/(r + R)^2

I began by multiplying both sides by (r +R)^2.

I then got this:

p(r + R)^2 = E^(2) R

I then simplified the left side and got this:

pr^2 + 2rR + R^2 = E^2 R

I subtracted pr^2 from both sides and got this:

2rR + R^2 = E^2 R - pr^2

I then subtracted both sides by E^2 R.

2rR + R^2 - E^2 R = -pr^2

I then factored out the upper case letter R on the left side and got this:

R(2r + R - E^2) = -pr^2

This is where I got stuck.

Notice that there is another upper case letter R inside the parentheses.

Where do I go from there?

2. Originally Posted by magentarita
p = [E^(2) R]/(r + R)^2

I began by multiplying both sides by (r +R)^2.

I then got this:

p(r + R)^2 = E^(2) R

I then simplified the left side and got this:

pr^2 + 2prR + pR^2 = E^2 R This is where you went wrong. You didn't multiply every term by p.

This is a monster. Where did this come from?

$\displaystyle pr^2+2prR+pR^2=E^2R$

$\displaystyle pr^2+2prR+pR^2-E^2R=0$

$\displaystyle pR^2+2prR-E^2R=-pr^2$

$\displaystyle pR^2+(2pr-E^2)R=-pr^2$

Divide everything by p.

$\displaystyle R^2+\left(\frac{2pr-E^2}{p}\right)R=-r^2$

Try completing the square.

$\displaystyle R^2+\left(\frac{2pr-E^2}{p}\right)R+\left(\frac{2pr-E^2}{2p}\right)^2=-r^2+\left(\frac{2pr-E^2}{2p}\right)^2$

$\displaystyle \left(R+\frac{2pr-E^2}{2p}\right)^2=-r^2+\left(\frac{2pr-E^2}{2p}\right)^2$

Take the square root of both sides.

$\displaystyle R+\frac{2pr-E}{2p}=\pm\sqrt{-r^2+\left(\frac{2pr-E}{2p}\right)^2}$

$\displaystyle R=\pm\sqrt{-r^2+\left(\frac{2pr-E}{2p}\right)^2}-\frac{2pr-E}{2p}$

I don't know if the right side gets any better. Maybe an algebraist will take a look at this and submit a cleaner solution.

3. ## thanks............

Originally Posted by masters
This is a monster. Where did this come from?

$\displaystyle pr^2+2prR+pR^2=E^2R$

$\displaystyle pr^2+2prR+pR^2-E^2R=0$

$\displaystyle pR^2+2prR-E^2R=-pr^2$

$\displaystyle pR^2+(2pr-E^2)R=-pr^2$

Divide everything by p.

$\displaystyle R^2+\left(\frac{2pr-E^2}{p}\right)R=-r^2$

Try completing the square.

$\displaystyle R^2+\left(\frac{2pr-E^2}{p}\right)R+\left(\frac{2pr-E^2}{2p}\right)^2=-r^2+\left(\frac{2pr-E^2}{2p}\right)^2$

$\displaystyle \left(R+\frac{2pr-E^2}{2p}\right)^2=-r^2+\left(\frac{2pr-E^2}{2p}\right)^2$

Take the square root of both sides.

$\displaystyle R+\frac{2pr-E}{2p}=\pm\sqrt{-r^2+\left(\frac{2pr-E}{2p}\right)^2}$

$\displaystyle R=\pm\sqrt{-r^2+\left(\frac{2pr-E}{2p}\right)^2}-\frac{2pr-E}{2p}$

I don't know if the right side gets any better. Maybe an algebraist will take a look at this and submit a cleaner solution.
Yes, it is a monster and it's great from start to finish.
Don't you just love math? I'll take a math question over date anytime.