Maths Difficulty.....Please Help Somebody :)

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    (Original post by theone)
    Sure, all quadratics can be written in the form x^2 +bx + c.

    Now we can note by multiplying out that this is in fact equal to (x+(b/2))^2 + (c - b^2/4).

    This procedure of getting the second equation from the first is called completing the square.

    By extending this you can work out the more general formula for the solution of a quadratic.
    Theone could you just confirm that:

    (x) = 3[(x-1r1r3)^2+10r2r3] in completed square form

    is the right answer.
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    just draw it and look
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    (Original post by clever_lad)
    i am correct, you dont ahve a clue what you are talking about
    No, you've forgotten to divide the 8x (the 4) by 3 when you took it out... it should be 4/3.
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    (Original post by Bhaal85)
    Theone could you just confirm that:

    (x) = 3[(x-1r1r3)^2+10r2r3] in completed square form

    is the right answer.
    Well, it ain't going to be since you have - 4/3 and the original equation has a positive term in x.

    I get f(x) = 3((x+4/3)^2 - 16/9) which gives x = -4/3 as the max.

    You get the same by differentiation
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    (Original post by ++Hex++)
    No, you've forgotten to divide the 8x (the 4) by 3 when you took it out... it should be 4/3.
    So I am right. And that arse was just being rude and arrogant?
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    (Original post by theone)
    Well, it ain't going to be since you have - 4/3 and the original equation has a positive term in x.

    I get f(x) = 3((x+4/3)^2 - 16/9) which gives x = -4/3 as the max.

    You get the same by differentiation
    *goes and gets pen and paper*
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    (Original post by Bhaal85)
    So I am right. And that arse was just being rude and arrogant?
    Who's the arse?
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    (Original post by Bhaal85)
    So I am right. And that arse was just being rude and arrogant?
    No need for that. Everyone makes mistakes. Especially where maths is involved...
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    Sorry i've made and error it should read

    3((x+4/3)^2-32/9) i think.
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    (Original post by theone)
    Who's the arse?
    notsoclever_lad

    Yeah your right dy/dx=0 at x=-8/6. ben revising for Discrete for 4 weeks. lol. Not arsed, but that arse was wrong.
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    (Original post by clever_lad)
    i am correct, you dont ahve a clue what you are talking about
    Thats what the arse wrote, to m that is being rude and arrogant, considering I was helping him.
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    i was right!!
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    (Original post by Bhaal85)
    Thats what the arse wrote, to m that is being rude and arrogant, considering I was helping him.
    Oh, ok, thought that you were having a go at me, no offence
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    (Original post by theone)
    Sorry i've made and error it should read

    3((x+4/3)^2-32/9) i think.
    10r2r3 not 32/9
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    (Original post by cimmie)
    i was right!!
    hmmpf whats about funny pictures??
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    (Original post by cimmie)
    i was right!!
    how are you right??? Hmm...you must be notsoclever_lad then???
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    (Original post by Bhaal85)
    10r2r3 not 32/9
    No i don't think.... since we want to complete the square of x^2 +8x/3 +16/3 = (x+4/3)^2 + 16/3 - 16/9.

    And 16/3 - 16/9 = 32/9
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    oops, sorry Bhall, i didnt complete the square correctly. i am really sorry mate, i take my comment back *shrieks with embarrasment*........if you dont mind mate, please could you go through the process of completing the square for this equation, then showing me what x value will give the minimum value for the parabola.....please mate...once again, sorry. *shudders*
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    (Original post by cimmie)
    x=-4/3

    f'(x):6x+8
    =>6x+8=0
    <=>x=-4/3

    sorry its just guessed
    i had it first...
    but forget this stupid thing! Its just confusing me!
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    (Original post by theone)
    No i don't think.... since we want to complete the square of x^2 +8x/3 +16/3 = (x+4/3)^2 + 16/3 - 16/9.

    And 16/3 - 16/9 = 32/9
    dy/dx = 6x+8 right.

    set to equal 0 = 6x+8=0
    rearrange to make x subject you get: x=-8/6

    sub that into orignal equation 3x^2+8x+16 = 10r2r3. (ten and two thirds)

 
 
 
Updated: October 28, 2003
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