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Math Help - Quadratic Equations

  1. #1
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    Quadratic Equations

    Hello there, I've got a quadratic equation. I'm supposed to solve it by factorising but it's already factorized.

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

    Can you help me?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Quadratic Equations

    Use the fact that a\cdot b=0 \Leftrightarrow a=0 \ \mbox{or} \ b=0
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  3. #3
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    Re: Quadratic Equations

    For (3x + 1)(x + 3) = 19 - 2(x +2)

    Do I expand everything on the left hand side then do the same on the the Right hand side then move the right hand side to the left hand side to make it = 0?
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  4. #4
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    Re: Quadratic Equations

    Yep have a go at that.
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  5. #5
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    Re: Quadratic Equations

    After a long complex situation I get

    x^2+10x+18=0

    Which I can't do as nothing multiplys to make 18 thats add to 10
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  6. #6
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    Re: Quadratic Equations

    For \displaystyle ax^2+bx+c=0\implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}
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  7. #7
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    Re: Quadratic Equations

    But I've got to do via factorisation. Don't think I can factorise the above. Think I done something wrong.
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  8. #8
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    Re: Quadratic Equations

    Complete the square then.
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  9. #9
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    Re: Quadratic Equations

    Quote Originally Posted by JibblyJabbly View Post
    But I've got to do via factorisation
    x^2 + 10x + 18 = 0

    if you insist ...

    (x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0
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  10. #10
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    Re: Quadratic Equations

    Quote Originally Posted by skeeter View Post
    x^2 + 10x + 18 = 0

    if you insist ...

    (x + 5 + \sqrt{7})(x + 5 - \sqrt{7}) = 0
    Just to double check
    (3x + 1)(x + 3) = 19 - 2(x +2) is equal to

    x^2 + 10x + 18 = 0 right? I assume I had problems with the Left hand side of the equation
    Last edited by JibblyJabbly; September 21st 2011 at 06:38 PM.
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  11. #11
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    Re: Quadratic Equations

    Better check my workings but,

    (3x + 1)(x + 3) = 19 - 2(x +2)

    3x^2+9x+x+3 = 19 - 2x -4

    3x^2+10x+3 = 15 - 2x

    3x^2+12x-12 = 0

    3(x^2+4x-4) = 0

    x^2+4x-4 = 0
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