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Thread: Quadratic Equation

  1. #1
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    Quadratic Equation


    5x2
    -3x+12=3(x+4)


    There's supposed to be two different possible solutions for X, please help me find them.
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  2. #2
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    Re: Quadratic Equation

    Quote Originally Posted by ceedee View Post

    5x2
    -3x+12=3(x+4)


    There's supposed to be two different possible solutions for X, please help me find them.
    $ \begin{align*}5x^2-3x+12&=3(x+4) \\&=3x+12\\5x^2-6x&=0\\\text{Now you solve} \end{align*}$
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  3. #3
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    Re: Quadratic Equation

    Yes I also get that far, but I don't get further...help please and can you explain your answer then also please?
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    Re: Quadratic Equation

    Quote Originally Posted by ceedee View Post
    Yes I also get that far, but I don't get further...help please and can you explain your answer then also please?
    $5x^2 - 6x = 0$

    $x(5x-6) = 0$

    can you finish it now?

    If the product of two numbers is zero what can you say about at least one of those numbers?
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  5. #5
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    Re: Quadratic Equation

    Thank you for the answer, is it X=2 ?
    Last edited by ceedee; Jan 27th 2018 at 09:52 AM.
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  6. #6
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    Re: Quadratic Equation

    $5x^2 - 6x = 0$


    $\large ax^2+bx+c=0\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

    $a=5,~b=-6,~\&~c=0$
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  7. #7
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    Re: Quadratic Equation

    Quote Originally Posted by ceedee View Post
    Thank you for the answer, is it X=2 ?
    no...

    $x(5x-6) = 0$

    in order for the product to equal zero at least one of the factors must equal zero.

    So either

    $x=0$

    or

    $5x-6=0 \rightarrow x = \dfrac 6 5$

    so $x = 0$ OR $x=\dfrac 6 5$
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  8. #8
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    Re: Quadratic Equation

    Thank you very much
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  9. #9
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    Re: Quadratic Equation

    Thank you!!
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