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Math Help - quadratic equations

  1. #1
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    quadratic equations

    hi, could anyone help me understand how to do quadratic equations. i've got the equation 3x^2 - 7x + 2 and it needs to be factorised. i was hoping someone could show me a systematic way of doing it. i read somewhere that you need to find 2 numbers that add together to make -7 and multiply together to make 2 but i don't think that applies here. the answer is (x - 2) (3x - 1) . could someone show me a simple way of figuring it out so i don't have to resort to trial and error.

    thanks, Mark
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by mark View Post
    hi, could anyone help me understand how to do quadratic equations. i've got the equation 3x^2 - 7x + 2 and it needs to be factorised. i was hoping someone could show me a systematic way of doing it. i read somewhere that you need to find 2 numbers that add together to make -7 and multiply together to make 2 but i don't think that applies here. the answer is (x - 2) (3x - 1) . could someone show me a simple way of figuring it out so i don't have to resort to trial and error.

    thanks, Mark
    Hi mark,

    Here's a technique you can use on this type of quadratic expression where the leading coefficient is not 1.

    3x^2-7x+2

    First, multiply the leading coefficient by the constant:

    3 \times 2 = 6

    Think of two numbers whose product is 6 and sum is -7.

    You should come up with -1 and -6 pretty quickly.

    Replace the middle term of your original quadratic with these two factors.

    3x^2-1x-6x+2

    Group the terms and factor:

    (3x-1x)-(6x-2)

    x(3x-1)-2(3x-1)

    (x-2)(3x-1)
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  3. #3
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    hi masters (or anyone reading this),

    i don't think i understand what happens when you get to group the terms and factor.

    (3x^2 - 1x) - (6x - 2) i understand that part just shown, but then why do you go on to x(3x - 1x) -2(3x - 1) ?

    and how do you get (x - 2) (3x - 1) after all that?

    thanks
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  4. #4
    Senior Member nikhil's Avatar
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    Lightbulb

    (3x^2-1x)-(6x-2)
    taking x common from (3x^2-1x) and 2 from (6x-2) we get
    [x(3x-1)-2(3x-1)]
    now from the whole term take (3x-1) common so we get
    (3x-1)(x-2)
    this is done for FACTORIZATION
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  5. #5
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    from the whole term take (3x-1) common? so you just remove it completely? why do you do that? (if thats what you do)
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  6. #6
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    You could simply re-expand the new term to see that it is the same:

     (3x-1)(x-2) = 3x^2 -7x +2
     x(3x-1) -2 (3x-1) = 3x^2 - x -6x + 2 = 3x^2 -7x +2
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  7. #7
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    thanks
    Last edited by mark; August 26th 2009 at 04:15 AM.
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