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Simple square root addition question

  1. Oct 4, 2008 #1
    I know this must be an easy question, but I can't seem to remember how to do it:

    [tex]\sqrt{z^2+R^2 - 2zR} - \sqrt{z^2+R^2 + 2zR}[/tex]

    Can someone go through step by step how to solve this? This isn't a homework question but I've been running into this problem more often in multiple courses.

    So far, I've gotten down to something like this but I'm not sure if this is the right track. I replaced parts of the above equation to make it simpler but I STILL can't remember how to do it:

    [tex]\sqrt{C^2 - x} - \sqrt{C^2 + x}[/tex]

    Thanks
     
  2. jcsd
  3. Oct 4, 2008 #2

    Doc Al

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    Expand these expressions:
    (A + B)² = ?
    (A - B)² = ?
     
  4. Oct 4, 2008 #3

    arildno

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    Simplify even further, and utilize the trick:
    [tex]a-b=(a-b)\frac{a+b}{a+b}=\frac{a^{2}-b^{2}}{a+b}[/tex]
    This type of simplification can be necessary prior to performing finite-digit arithmetic on the computation (i.e, by a computer), since otherwise, there might result a loss of significant digits when subtracting almost equally large numbers from each other.
     
  5. Oct 4, 2008 #4

    arildno

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    Oops, I forgot those two..:shy:
     
  6. Oct 4, 2008 #5
    Doc Al, thanks for the hint! I was able to figure it out now.

    I do have another similar problem in another course with unequal variables. For instance, I have to calculate the momentum of some subatomic particles and I get an equation like:

    1000 =[tex]\sqrt{(pc)^2 + 875000} + \sqrt{(pc)^2 + 130000}[/tex]

    How would I solve for "pc" in this case?
     
  7. Oct 4, 2008 #6

    Doc Al

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    No easy trick here, I'm afraid. Just get rid of those square roots. Call (pc)² = x (to keep it simple) and square both sides. Then rearrange and do it again. (You'll end up with a quadratic to solve.)
     
  8. Oct 4, 2008 #7

    Hootenanny

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    :surprised What are you doing in Maths, Doc?
     
  9. Oct 4, 2008 #8

    Doc Al

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    I took a wrong turn somewhere... :uhh:
     
  10. Oct 4, 2008 #9
    Thanks Doc Al and arildno for the quick responses.

    Do you know what the solution is for "pc" in that example above? I am still in the process of trying to solve that one but my skills are a little rusty.

    So if you square both sides, do you get:

    [tex]1000^2 = [\sqrt{(pc)^2 + 875000} + \sqrt{(pc)^2 + 130000} ]^2[/tex]

    I am having trouble in trying to simplify the right side.
     
  11. Oct 4, 2008 #10

    Doc Al

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    Use this "trick" again: (A + B)² = ?
     
  12. Oct 4, 2008 #11
    Super! It took awhile but at least I got this down pretty good again. I did some searching online to refresh my square root concepts and solve the rest. I solved a problem in the book that I knew the answer to, but the problem above is one that I just made up randomly so I didn't know the answer.

    I got pc = 337 approximately. If you solved this, let me know if you got the same thing.

    Thanks
    Regards
     
  13. Oct 4, 2008 #12

    Doc Al

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    I'm glad that this is just a made up example, since it has no real solution (only an imaginary one)! (I didn't realize that until I tried to solve it myself.)

    Pick a different example to test your skills:

    1000 = √[(pc)^2 + 1000] + √[(pc)^2 + 2000]
     
  14. Oct 4, 2008 #13
    I was able to get about 498.5 roughly. Is this correct?
     
  15. Oct 4, 2008 #14
    Actually, all I would just have to do is plug that answer back into the equation and it looks like it is correct.

    Thanks for the help Doc Al
     
  16. Oct 4, 2008 #15

    Doc Al

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    Looks good.
    Now you're thinking. :approve:
    You are welcome.
     
  17. Oct 4, 2008 #16

    Redbelly98

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    If I substitute pc=337 into the equation of post #5, I don't get 1000.

    Try this: what value of pc gives a minimum for the right-hand-side expression? How does that value compare with the 1000 on the left-hand-side?
     
  18. Oct 4, 2008 #17

    Doc Al

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    Try substituting 337i. :wink:
     
  19. Oct 4, 2008 #18

    Redbelly98

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    Much better. I will sleep well tonight.
     
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