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Math Help - Finding the x- intercept

  1. #1
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    Finding the x- intercept

    I would just like a second opinion please to ensure I am doing this correctly.

    When y = 0

    0 = -x^2 + 4x + 5
    - 5 = 4x

    x = - 1.25
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  2. #2
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    Re: Finding the x- intercept

    Quote Originally Posted by David Green View Post
    I would just like a second opinion please to ensure I am doing this correctly.

    When y = 0

    0 = -x^2 + 4x + 5
    - 5 = 4x

    x = - 1.25
    The x^2 term does not disappear.

    \displaystyle \begin{align*} 0 &= -x^2 + 4x + 5 \\ x^2 - 4x - 5 &= 0 \\ (x + 1)(x - 5) &= 0 \\ x + 1 = 0 \textrm{ or } x - 5 &= 0 \\ x = -1 \textrm{ or } x &= 5 \end{align*}
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Finding the x- intercept

    What happened with the -x^2?
    Use the quadratic formula.
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  4. #4
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    Re: Finding the x- intercept

    Quote Originally Posted by Siron View Post
    What happened with the -x^2?
    Use the quadratic formula.
    Not sure if this works?

    - b + or - square root b^2 - 4ac / 2a

    b = 4, a = -1, and c = 5

    4^2 - 4 x 1 x 5 = 16 - 20 = - 4

    square root of - 4 = E?
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  5. #5
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    Re: Finding the x- intercept

    Quote Originally Posted by David Green View Post
    Not sure if this works?

    - b + or - square root b^2 - 4ac / 2a

    b = 4, a = -1, and c = 5

    4^2 - 4 x 1 x 5 = 16 - 20 = - 4

    square root of - 4 = E?
    The discriminant is \displaystyle 4^2 - 4(-1)(5) = 16 + 20 = 36, so there are two solutions.

    Of course, the easier method is to look at my post above and realise that the quadratic easily factorises.
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  6. #6
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    Re: Finding the x- intercept

    Quote Originally Posted by Prove It View Post
    The discriminant is \displaystyle 4^2 - 4(-1)(5) = 16 + 20 = 36, so there are two solutions.

    Of course, the easier method is to look at my post above and realise that the quadratic easily factorises.
    I do keep getting my minus and a minus confused?

    Thanks
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Finding the x- intercept

    In the second degree equation:
    -x^2+4x+5=0
    is a=(-1), b=4, c=5 that means if we substitute this values in the formula D=b^2-4ac we get:
    D=(4)^2-4(-1)(5)=36
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  8. #8
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    Re: Finding the x- intercept

    Quote Originally Posted by Siron View Post
    In the second degree equation:
    -x^2+4x+5=0
    is a=(-1), b=4, c=5 that means if we substitute this values in the formula D=b^2-4ac we get:
    D=(4)^2-4(-1)(5)=36
    Interestingly I still get confused with my rules?

    36 squared = 6 / 2 = 3, 3 + (-4) = -1 first root

    36 squared = 6 / 2 = 3, 3 - ( - 4) = 7 ?

    How do I get the root 5?
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  9. #9
    MHF Contributor Siron's Avatar
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    Re: Finding the x- intercept

    There're two solutions (if D\geq0) for a quadratic equation by using the quadratic formula:
    x_1,x_2=\frac{-b\pm\sqrt{D}}{2a}
    So in this case:
    x_1=\frac{-4+6}{-2}=-1
    x_2=\frac{-4-6}{-2}=\frac{-10}{-2}=5
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  10. #10
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    Re: Finding the x- intercept

    Quote Originally Posted by Siron View Post
    There're two solutions (if D\geq0) for a quadratic equation by using the quadratic formula:
    x_1,x_2=\frac{-b\pm\sqrt{D}}{2a}
    So in this case:
    x_1=\frac{-4+6}{-2}=-1
    x_2=\frac{-4-6}{-2}=\frac{-10}{-2}=5
    I can't say that I can follow the reasoning there yet, what happend to the rest of the formula?
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  11. #11
    MHF Contributor Siron's Avatar
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    Re: Finding the x- intercept

    What do you mean with the 'rest of the formula'? Don't you understand how I got the solutions? ...
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  12. #12
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    Re: Finding the x- intercept

    I understand the two roots and how you got them, but the formula has been modified.

    - b + or - square root D / 2a

    See the changes I have not seen before?

    Capital D not squared, and - 4ac missing?

    Like I said before this subject is new to me and is a new learning curve.
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  13. #13
    MHF Contributor Siron's Avatar
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    Re: Finding the x- intercept

    I don't think the formula has been modified.
    In general if you have a quadratic equation:
    ax^2+bx+c=0
    Then this equation has:
    two different solutions (or roots) if D=b^2-4ac>0
    x_1=\frac{-b+\sqrt{D}}{2a}
    x_2=\frac{-b-\sqrt{D}}{2a}

    If D=0 then the two identical solutions are:
    x_1,x_2=\frac{-b}{2a}

    If D<0:
    There're no real solutions.

    Is this clear? If you need a prove of this then you can ask it.
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  14. #14
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    Re: Finding the x- intercept

    So the capital D, this I now understand to be the solution of;

    b^2 - 4ac

    then the answer = D which I take the square root value.
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  15. #15
    MHF Contributor Siron's Avatar
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    Re: Finding the x- intercept

    D is the discriminant, and indeed D=b^2-4ac.

    I think you get the point now. Do you?
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