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Sum of two numbers

  1. Feb 28, 2008 #1
    The sume of two numbers is 50, and their product is 900. Find the exact value of the sum of their reciprocals.

    [tex]x+y=50[/tex]

    [tex]xy=900[/tex]

    [tex]x^2-50x+900[/tex]

    Imaginary solutions?
     
  2. jcsd
  3. Feb 28, 2008 #2

    HallsofIvy

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    Blast!! I just spent 10 minutes writing out an explanation of why "if x+y= 50, xy cannot be more than 225" when I suddenly noticed the word "reciprocals"!!!

    x+ y= 50 and 1/x+ 1/y= 900.
     
  4. Feb 28, 2008 #3
    did u mean here xy cannot be more than 625, because let x=y=25, then
    xy=25*25=625
     
  5. Feb 28, 2008 #4
    to the op
    like halls started
    1/x+1/y=(x+y)/xy =50/900
     
  6. Feb 28, 2008 #5
    i do not think this part is correct!!
     
  7. Feb 28, 2008 #6

    symbolipoint

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    Yes, imaginary solutions. Starting from the form, [tex] \[
    x^2 - 50x + 900 = 0
    \]
    [/tex], and directly using general quadratic equation solution, part of the expression will contain [tex] \[
    \sqrt {2500 - 3600}
    \]
    [/tex], which is imaginary.
     
    Last edited: Feb 28, 2008
  8. Feb 28, 2008 #7
    well then, can't one do what i did?

    let 1/x be the reciprocal for x, and 1/y the reciprocal of y, so we want to find the sum of them, and we have:

    1/x+1/y=(x+y)/xy=50/900=5/90 , what is wrong with this???
     
  9. Feb 28, 2008 #8

    symbolipoint

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    Not sure, since I did not carry the solution far enough. I only developed a solution for x and y. Try taking those solutions and then yourself carry farther to the extent of the question asked (about the sum of the reciprocals) and find out how it works. I did not disagree with you; I merely did not take the intermediate solutions far enough for the original question.
     
  10. Feb 28, 2008 #9

    Dick

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    That's the clever way to do it. If you find the roots like, x=25+5*sqrt(11)*i and y=conjugate(x), invert them and add them, you'll get... 50/900.
     
  11. Feb 28, 2008 #10

    HallsofIvy

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    Well, I did a bit more than just calculate that value. Assuming, as was implied by the work but is wrong, that the equations are x+ y= 50 and xy= 900, replace y with 50-x so x(50- x)= 900. The left side is a quadratic function. Completing the square, I found that the vertex of its graph is at (25, 625) and so the value on the left side cannot be more than 625.
     
  12. Feb 28, 2008 #11

    HallsofIvy

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    Right. I got so annoyed at doing useless work I just wrote it out wrong. Your solution is very nice.
     
  13. Feb 28, 2008 #12
    Yeah, but you wrote there "cannot be more than 225", and i know this was a typo, so i just wanted to point it out!
     
  14. Feb 28, 2008 #13
    Wow, very nice! Thanks :-]
     
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