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Math Help - x intercept of parabola

  1. #1
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    x intercept of parabola

    Just a bit confused here and would appreciate a little guidance please.

    Find the x intercept when y=0

    0 = -x^2 + 4x +5

    x^2 - 4x - 5 = 0

    (x-1)(x-5) = 0 (Something here I think is not quite right)?

    multiply out the brackets;

    x^2 - 5x - x +5 = 0

    x^2 - 6x + 5 = 0 (This can't be right it does'nt equal the original equation)?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: x intercept of parabola

    Note: Recalculate the discriminant. It has to be:
    -x^2+4x+5=(x+1)\cdot (x-5)
    Because:
    (x+1)\cdot (x-5)=x^2-5x+x-5=x^2-4x-5
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  3. #3
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    Re: x intercept of parabola

    Thanks for that Siron, I see where I was going wrong now, I said (x - 1) you pointed out (x + 1)

    So now it works

    Thanks
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: x intercept of parabola

    You're welcome! In this case using the discriminant is not necessary, you can just factore it directly. But if you want, you'll see -1 is a zero in stead of 1.
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  5. #5
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    Re: x intercept of parabola

    Quote Originally Posted by Siron View Post
    Note: Recalculate the discriminant. It has to be:
    -x^2+4x+5=(x+1)\cdot (x-5)
    Because:
    (x+1)\cdot (x-5)=x^2-5x+x-5=x^2-4x-5
    This I think I found a problem?

    (x + 1)(x - 5) = x^2 -5x + x - 5

    = x^2 -4x - 5 ?

    should minus 5 not be +5?

    (x - 1)(x + 5) = x^2 +5x - x + 5

    = x^2 + 4x + 5

    Sorry you were right before I change things, if I do this;

    - x^2 - 4x - 5 = 0

    It's wrong.
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: x intercept of parabola

    But (+1)(-5)=-5 and (-1)(+5)=-5
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  7. #7
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    Re: x intercept of parabola

    -x^2 + 4x + 5 = 0

    -(x^2 - 4x - 5) = 0

    x^2 - 4x - 5 = 0

    (x-5)(x+1) = 0

    x = 5 , x = -1
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