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Math Help - 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 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 -7x into -8x + x

    Now you have

    2x^2 - 7x - 4

     = 2x^2 - 8x + x - 4

     = 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...

    2x^2 - 7x - 4 = 0

    (2x + 1)(x - 4) = 0

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

    So x = -\frac{1}{2} or 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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