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Math Help - Quadratics: Completing the square

  1. #1
    Junior Member Turple's Avatar
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    Exclamation Quadratics: Completing the square

    Hi, I've been doing some GCSE maths revision - working on completing the square. I was going through questions like x+4x-3=0 using (x+a)+b and then solving them. However, I reached questions with numbers in front of the x and couldn't understand how to solve it:

    2x-6x+1=0

    I was wondering if somebody could please help, thankyou.
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  2. #2
    Moo
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    Hello,

    Factorise by 2 and then do as usual
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  3. #3
    Junior Member Turple's Avatar
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    I don't understand - how can I factorise it by 2 when there are no numbers that multiply to give one and add to give -6? Sorry!
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by Turple View Post
    Hi, I've been doing some GCSE maths revision - working on completing the square. I was going through questions like x+4x-3=0 using (x+a)+b and then solving them. However, I reached questions with numbers in front of the x and couldn't understand how to solve it:

    2x-6x+1=0

    I was wondering if somebody could please help, thankyou.
    2x^2-6x+1=0

    Let's take that two out.

    2(x^2 - 3x) + 1 = 0

    Complete the square.

    In this bracket, (x^2 - 3x) , we want to add +2,25 inside there.
    Why? Because the co-efficient of the -3x could only have been possible when we had 2 \times -1,5.
    And: (-1,5)^2 = +2,25

    But we cant forget about the 2 outside the bracket. 2(x^2 - 3x)

    So now we can say this:

    2(x^2 - 3x) + 2(2,25) - 2(2,25) + 1 = 0

    2(x^2 - 3x + 2,25) - 4,5 + 1 = 0

    2(x - 1,5)^2 - 3,5 = 0
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  5. #5
    Moo
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    Oh i see, sorry ^^

    2x^2-6x+1=2(x^2-3x+\frac{1}{2})=2(x^2-2*\frac{3}{2} x+\frac{1}{2})

    (\frac{3}{2})^2=\frac{9}{4}

    2(\underbrace{x^2-2*\frac{3}{2} x+\frac{9}{4}}_{(x-\frac{3}{2})^2}-\frac{9}{4}+\frac{1}{2})=2(x-\frac{3}{2})^2+2(\underbrace{-\frac{9}{4}+\frac{1}{2}}_{-\frac{7}{4}})

    So 2x^2-6x+1=2(x-\frac{3}{2})^2-\frac{7}{2}

    I haven't checked my calculus yet, but did you understand how to do ?
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by Moo View Post
    Oh i see, sorry ^^

    2x^2-6x+1=2(x^2-3x+\frac{1}{2})=2(x^2-2*\frac{3}{2} x+\frac{1}{2})

    (\frac{3}{2})^2=\frac{9}{4}

    2(\underbrace{x^2-2*\frac{3}{2} x+\frac{9}{4}}_{(x-\frac{3}{2})^2}-\frac{9}{4}+\frac{1}{2})=2(x-\frac{3}{2})^2+2(\frac{-\frac{9}{4}+\frac{1}{2}}_{-\frac{7}{4}})

    So 2x^2-6x+1=2(x-\frac{3}{2})^2-\frac{7}{2}

    I haven't checked my calculus yet, but did you understand how to do ?
    Beat you to it, Moo
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  7. #7
    Moo
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    I hate writing in latex

    I'll get my revenge jan
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  8. #8
    Junior Member Turple's Avatar
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    Woah. Sorry but I am so confused XD I can't understand where the 2.25 comes from?
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  9. #9
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Turple View Post
    Woah. Sorry but I am so confused XD I can't understand where the 2.25 comes from?
    2.25 = 1.5^2
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  10. #10
    Junior Member Turple's Avatar
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    So where does that 1.5 come from?

    Sorry :/
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  11. #11
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Turple View Post
    So where does that 1.5 come from?

    Sorry :/
    (x - 1.5)^2 = x^2 - 3x + 2.25 (Eugh, I hate typing a dot for the decimal seperator)

    Do you see now?
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  12. #12
    Moo
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    See my demo, there are fractions and no questioning about "where does it come"

    Ok ok i'm leaving ^^
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  13. #13
    Junior Member Turple's Avatar
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    So there is an x and a -3x? Has the orginial quadratic been halved? And if it has I'm lost with that 2.25 again...
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  14. #14
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Moo View Post
    Ok ok i'm leaving ^^
    I'm thinking you should rather take over...
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  15. #15
    Junior Member Turple's Avatar
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    Ok Moo I will try to understand yours XD

    Sorry to both fo you if I'm being frustratingly thick :/
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