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Math Help - Unsure of how to solve this right triangle

  1. #1
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    Unsure of how to solve this right triangle

    Unsure of how to solve this right triangle-bc810cfff18d0391268226fb9d4e5f.gif
    Thats 29 squared on the hypotenuse and x+3 on the opposite and just x on the adjacent. I cant figure what to do with the 29 after it is given an exponent of 2 and taken out of square form.
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  2. #2
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    Re: Unsure of how to solve this right triangle

    Quote Originally Posted by goldbug78 View Post
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    Thats 29 squared on the hypotenuse and x+3 on the opposite and just x on the adjacent. I cant figure what to do with the 29 after it is given an exponent of 2 and taken out of square form.

    You need to solve x^2+(x+3)^2=29.
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    Re: Unsure of how to solve this right triangle

    yeah, but what do with 3x+6x+9=29?I am not sure how to solve this quadratic expression. I know 3x is raised to the second power, just can't figure how to use the tools on this site yet.
    Last edited by goldbug78; January 22nd 2013 at 11:08 AM.
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    Re: Unsure of how to solve this right triangle

    Quote Originally Posted by goldbug78 View Post
    yeah, but what do with 3x+6x+9=29?I am not sure how to solve this quadratic expression. I know 3x is squared, just can't figure how to use the tools on this site yet.
    \\x^2+(x+3)^2=29\\2x^2+6x+9=29\\2x^2+6x-20=0
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  5. #5
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    Re: Unsure of how to solve this right triangle

    I got it now, with the answer 2. I kept thinking it was 3x instead of 2x. Thanks for the help!
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    Re: Unsure of how to solve this right triangle

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  7. #7
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    Re: Unsure of how to solve this right triangle

    x + (x + 3) = (√29)
    x + x + 6x + 9 = 29
    2x + 6x 20 = 0
    x + 3x 10 = 0
    x = [ 3 √(3 + 40)]/2
    = [ 3 √49]/2
    = ( 3 7)/2
    = taking + ve values
    x = ( 3 + 7)/2 = 2
    sides are 2, 5, √29 units
    ----
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