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Math Help - Quadratic Problem,

  1. #1
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    Quadratic Problem,

    I've been trying to figure out this question but i dont know where im going wrong when i plug back in the solutions i get, i get a bunch of wacky numbers...

    question : x^2 + (x + 2) ^2 = 290

    my work:
    = x^2 + (x +2) + (x + 2) = 290
    = 2x^2 + 4x + 4 = 290
    = 2x + 4x - 286 = 0
    = 2 (x^2 + 2x - 143) = 0
    = (x - 11) ( x + 13)
    x = 11
    x = -13 .. why is this wrong when i check?
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  2. #2
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    Well, the first thing you have done wrong is (x+2)^2 does not equal to (x+2)+(x+2), it's (x+2)*(x+2)
    so x^2+(x+2)*(x+2)=290
    x^2+x^2+4x+4=290
    2x^2+4x+4=290
    2x^2+4x-286=0
    now, you can either solve this by the quadradic formula, or factoring
    QF: x=[-4√(16+2288)]/4
    x=(-448)/4
    x=11 or -13
    -------------------------------------
    on your work
    if you did x^2 + (x +2) + (x + 2) = 290, it should be x^2+2x+4, how did you get 2x^2 + 4x + 4 = 290?
    -------------------------------------
    Check: plug in 11
    121+169=290
    169+121=290
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  3. #3
    Senior Member
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     121 + (13)^2 = 290

     121+ 169 = 290

     290 = 290

    NOthing wrong with it...

     169 + 121 = 290


    You're correct spadda
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  4. #4
    Super Member Quacky's Avatar
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    Windsor, South-East England
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    Quote Originally Posted by jgv115 View Post
     121 + (13)^2 = 290

     121+ 169 = 290

     290 = 290

    NOthing wrong with it...

     169 + 121 = 290


    You're correct spadda
    He may be correct, but his notation is still blatantly misleading.
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