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Thread: Factorise and solve...

  1. #1
    Member Mr Rayon's Avatar
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    Smile Factorise and solve...

    Factorise and solve:

    2x^2 - 7x - 4

    Also, how do you factorise using cross multiplication? Somebody told me to do that but I didn't know what that meant. Anyway...if anyone could show me that would be great also!
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  2. #2
    Member Mr Rayon's Avatar
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    Sorry...I meant:

    2x^2 - 7x - 4 = 0


    ...I forgot how to edit a post.
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  3. #3
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    Quote Originally Posted by Mr Rayon View Post
    Factorise and solve:

    2x^2 - 7x - 4

    Also, how do you factorise using cross multiplication? Somebody told me to do that but I didn't know what that meant. Anyway...if anyone could show me that would be great also!
    You don't need to use the cross method. I hate it and the way I do it is more systematic.

    Multiply your a and c values. In this case it would be $\displaystyle 2\times -4 = -8$

    Now find 2 numbers that multiply to become the number you've just found (ac) and add to become b.

    So you need two numbers that multiply to become -8 and add to become -7. They are -8 and +1.

    So break up $\displaystyle -7x$ into $\displaystyle -8x + x$

    Now you have

    $\displaystyle 2x^2 - 7x - 4$

    $\displaystyle = 2x^2 - 8x + x - 4$

    $\displaystyle = 2x(x - 4) + 1(x - 4)$

    [tex] = (2x + 1)(x - 4).


    I assume if you're trying to solve the original quadratic that it was set equal to 0...

    $\displaystyle 2x^2 - 7x - 4 = 0$

    $\displaystyle (2x + 1)(x - 4) = 0$

    By the null factor law, $\displaystyle 2x + 1 = 0$ or $\displaystyle x - 4 = 0$.

    So $\displaystyle x = -\frac{1}{2}$ or $\displaystyle x = 4$.
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  4. #4
    Member Mr Rayon's Avatar
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    Smile

    Thanks...that method seems easier than cross multiplication!
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