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Math Help - The Quadratic Formula.

  1. #1
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    Exclamation The Quadratic Formula.

    Hey there. I am having difficulties with the Quadratic formula,
    I know what goes where in the cases, but its just solving from there.
    like in this equasion.



    2x^2-5x=0



    I know how to get it into the quadratic formula
    this is what I got.
    Attached Thumbnails Attached Thumbnails The Quadratic Formula.-equasion.jpg  
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  2. #2
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    Quote Originally Posted by facesonfilm View Post
    Hey there. I am having difficulties with the Quadratic formula,
    I know what goes where in the cases, but its just solving from there.
    like in this equasion.



    2x^2-5x=0



    I know how to get it into the quadratic formula
    this is what I got.
    that is wrong. c is zero here
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  3. #3
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    I know C is zero there.. but isnt it in the form of Ax^2 + Bx + C?

    so then when you plug it into the quadratic formula, you dont input a C.
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    Quote Originally Posted by facesonfilm View Post
    I know C is zero there.. but isnt it in the form of Ax^2 + Bx + C?

    so then when you plug it into the quadratic formula, you dont input a C.
    you would not have the -4(2) under the square root. since that is -4ac and c is zero
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  5. #5
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    so would the c (which is zero) make the whole 4ac equal zero?
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    Quote Originally Posted by facesonfilm View Post
    so would the c (which is zero) make the whole 4ac equal zero?
    umm, yes. zero times any number is zero
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  7. #7
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    Oh kay, I didnt even catch that. Wow.
    that sure makes me feel like a genius.
    thanks alot!!
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  8. #8
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    I'm surprised no one has pointed out that B is negative meaning that it should be

    \frac{5 \pm \sqrt{25}}{4}

    Since the quadratic is:

    \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

    Making it

    -(-5) or just  5
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    Quote Originally Posted by SnipedYou View Post
    I'm surprised no one has pointed out that B is negative meaning that it should be

    \frac{5 \pm \sqrt{25}}{4}
    we would have gotten to that

    we just had to deal with a more fundamental error first. anyway, so that we are all clear and on the same page:

    the formula is: x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}

    here a = 2, b = -5 and c = 0
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  10. #10
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    So because I have a negitive with a negitive, it becomes a positive,
    even in this case? Correct?
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  11. #11
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    Quote Originally Posted by facesonfilm View Post
    So because I have a negitive with a negitive, it becomes a positive,
    even in this case? Correct?
    yes, the negative of a negative is a positive
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  12. #12
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    okay, I wasn't 100 % on that because our teacher just gave us the formula today like 5 minutes before the bell so I was kinda foggy on that.
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  13. #13
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    why dont you just factorise "2x^2 -5x = 0"

    to "x(2x-5) = 0"

    then x would have the values:

    x = 0 and 2x-5 = 0
    2x = 5
    x =5/2

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  14. #14
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    Quote Originally Posted by shuggyboi View Post
    why dont you just factorise "2x^2 -5x = 0"

    to "x(2x-5) = 0"

    then x would have the values:

    x = 0 and 2x-5 = 0
    2x = 5
    x =5/2

    i think he was required to use the quadratic formula
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