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Math Help - Solve for R

  1. #1
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    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?

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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    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?

    pr^2+2prR+pR^2=E^2R

    pr^2+2prR+pR^2-E^2R=0

    pR^2+2prR-E^2R=-pr^2

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

    Divide everything by p.

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

    Try completing the square.

    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

    \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.

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

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

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

    pr^2+2prR+pR^2=E^2R

    pr^2+2prR+pR^2-E^2R=0

    pR^2+2prR-E^2R=-pr^2

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

    Divide everything by p.

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

    Try completing the square.

    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

    \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.

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

    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.
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