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Math Help - Radical Pythagorean Theorem

  1. #1
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    Radical Pythagorean Theorem

    I'm stumped on this.
    New problem!

    Original:
    x^2 + (x+5)^2 =(5√5)^2

    Now, I tried canceling out the radical. That gives me "0" for C when factoring, though, so it doesn't work out ( I believe).

    Doing 5^2 * 5 gives me 625, which gives me x=-20, 15 after factoring. Doesn't check out.
    Last edited by tyler.; April 13th 2009 at 04:36 PM.
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  2. #2
    Super Member Aryth's Avatar
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    If you notice:

    (5\sqrt{5})^2 = 5^2(\sqrt{5})^2 = 25*5 = 125

    So, now you have:

    x^2 + (x + 5)^2 = 125

    2x^2 + 10x + 25 = 125

    2x^2 + 10x = 100

    2x^2 + 10x - 100 = 0

    x^2 + 5x - 50 = 0

    Just solve that and you should have your answer.
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by tyler. View Post
    i'm stumped on this.
    New problem!

    Original:
    X^2 + (x+5)^2 =(5√5)^2

    now, i tried canceling out the radical. That gives me "0" for c when factoring, though, so it doesn't work out ( i believe).

    Doing 5^2 * 5 gives me 625, which gives me x=-20, 15 after factoring. Doesn't check out.
    5^2 * 5 = 5^3 = 125

    x^2 + x^2 + 10x + 25 = 125

    2x^2 + 10x - 100 = 0

    x^2 + 5x - 50 = 0

    and solve
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  4. #4
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    D'oh! I tried thinking that you should distrubute the square - but to the radicand even after simplifying. That was my big problem!

    Thanks a ton! This has been making me want to cry =P
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