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Math Help - Why is this incorrect (quadratics)?

  1. #1
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    Why is this incorrect (quadratics)?

    I'm supposed to solve:
    3X^2+17X+20=0

    This is my working:
    -> 3X^2-3X+20X+20=0
    -> 3X(X-1)-20(X-1)=0
    -> (X-1)(3X-20)=0
    (X-1) -> X=1
    (3X-20) -> 3X=20 -> X=20/3
    X= 1 or 20/3

    However, the anser to that question is
    -4 or -5/3

    What did I do wrong?
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  2. #2
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    Quote Originally Posted by juliak View Post
    I'm supposed to solve:
    3X^2+17X+20=0

    This is my working:
    -> 3X^2-3X+20X+20=0
    -> 3X(X-1)-20(X-1)=0 This is wrong. Should be 3x(x-1) + 20(x+1) which doesn't help.
    -> (X-1)(3X-20)=0
    (X-1) -> X=1
    (3X-20) -> 3X=20 -> X=20/3
    X= 1 or 20/3

    However, the anser to that question is
    -4 or -5/3

    What did I do wrong?
    In general, ax^2 + bx + c can be factored by finding two numbers that multiply to get ac and add up to b, call them r_{1}, r_{2}. Then you split your quadratic: ax^2 + r_{1}x + r_{2}x + c and you should be able to factor from there (assuming it was factorable in the first place)

    For your quadratic, you have to find 2 numbers that multiply to (3)(20) = 60 that add up to 17. 12 and 5 should come to mind. So:
    3x^2 + {\color{blue}17x} + 20 \ = \ 3x^2 + {\color{blue}12x + 5x} + 20 \ = \ 3x(x + 4) + 5(x+4) \ = \ \hdots
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  3. #3
    Super Member Matt Westwood's Avatar
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    What's with all the negative signs already?

    3x^2 + 17x + 20 = 0

    3x^2 - 3x + 20x + 20 = 0

    =3x(x-1) + 20 (x+1) = 0

    ... and your factorisation falls over.

    What I do in this situation when I can't find the numbers to factor by is use the quadratic formula. Then when I know what the answer is (or I've looked it up in the answers in the back of the book) I go back and use those numbers to factor it properly and I don't tell the teacher I cheated.
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  4. #4
    Super Member 11rdc11's Avatar
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    3x^2 + 17x+20

    Use the quad formula

    \frac{-17 ^+_-\sqrt{(17)^2-4(3)(20)}}{2(3)} = 0
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  5. #5
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    Oh okay I get it now.
    Heh, thank you .
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