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why do i need both positive and negative values (completing the square) Watch

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    (Original post by Zacken)
    I would expand the whole (x-2)^2 out to get: 3\bigg[x^2 -8x  + \frac{29}{3}\bigg] then you'd be halving -8 to get 3\bigg[(x-4)^2 + \cdots \bigg]
    (Original post by SeanFM)
    Neither. To me, it looks like you need to expand the brackets (x-2)^2 first.
    oh no this is part of a question sorry, the original question is express 3x^2 -12x + 17 in the form a(x+b)^2 + c

    i've got to this stage 3[(x-2)^2 - 4x + 17/3] but im unsure whether to half 4 or -4

    what i understand is i got to make it into this form (x±b/2)^2 - (b/2)^2 + C
    the b in (b/2)^2 confuses me because im not sure if b in this case is -4 or 4
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    (Original post by Zacken)
    I would expand the whole (x-2)^2 out to get: 3\bigg[x^2 -8x  + \frac{29}{3}\bigg] then you'd be halving -8 to get 3\bigg[(x-4)^2 + \cdots \bigg]
    (Original post by SeanFM)
    Neither. To me, it looks like you need to expand the brackets (x-2)^2 first.
    actually im fine! thank you i just figured it out myself
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    (Original post by ihatePE)
    oh no this is part of a question sorry, the original question is express 3x^2 -12x + 17 in the form a(x+b)^2 + c

    i've got to this stage 3[(x-2)^2 - 4x + 17/3] but im unsure whether to half 4 or -4

    what i understand is i got to make it into this form (x±b/2)^2 - (b/2)^2 + C
    the b in (b/2)^2 confuses me because im not sure if b in this case is -4 or 4
    The 'b' is always the number in front of the +x. So if you have -x, this is regarded as +(-1)(x) so b=-1. Also, I think this is a good sign for you to always include the original question in the future.
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    (Original post by Zacken)
    The 'b' is always the number in front of the +x. So if you have -x, this is regarded as +(-1)(x) so b=-1. Also, I think this is a good sign for you to always include the original question in the future.
    i know, i will do that from now on, also i need to know how to type mathsy signs on here properly, i can imagine its an eye sore for everyone the way i lay it out
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    (Original post by ihatePE)
    i know, i will do that from now on, also i need to know how to type mathsy signs on here properly, i can imagine its an eye sore for everyone the way i lay it out
    No, not really. The way you lay it out is fine.
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    tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

    Feels bad man.
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    (Original post by 34908seikj)
    tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

    Feels bad man.
    If it helps, it's 10 to 3 (a.m) here.
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    (Original post by 34908seikj)
    tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

    Feels bad man.
    I prefer completing the triangle at 2am.
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    (Original post by Zacken)
    I would expand the whole (x-2)^2 out to get: 3\bigg[x^2 -8x  + \frac{29}{3}\bigg] then you'd be halving -8 to get 3\bigg[(x-4)^2 + \cdots \bigg]
    HAI!!! just a petty question but in an expression such as 4(2-x)(2-x) how would i FOIL that because i dont know whether the 4 belongs to the first bracket or both brackets? thanks!
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    (Original post by 34908seikj)
    tfw it's 10 to 12 and you complete the square, fighting to keep your eyes open, only to find out OP no longer needs help

    Feels bad man.
    :rofl:
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    (Original post by ihatePE)
    HAI!!! just a petty question but in an expression such as 4(2-x)(2-x) how would i FOIL that because i dont know whether the 4 belongs to the first bracket or both brackets? thanks!
    It can belong to either one, i.e:

    4(2-x)(2-x) = (8-4x)(2-x) = (2-x)(8-4x) = 4(2-x)(2-x).

    I would personally leave the 4 outside for the moment, like so:

    \displaystyle 4\bigg[(2-x)(2-x) \bigg] = 4\bigg[4 - 4x + x^2\bigg] = 16 - 16x + 4x^2
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    (Original post by Zacken)
    It can belong to either one, i.e:

    4(2-x)(2-x) = (8-4x)(2-x) = (2-x)(8-4x) = 4(2-x)(2-x).

    I would personally leave the 4 outside for the moment, like so:

    \displaystyle 4\bigg[(2-x)(2-x) \bigg] = 4\bigg[4 - 4x + x^2\bigg] = 16 - 16x + 4x^2
    can you leave it out like what you've done for all expressions like 4(2-x)(2-x) ?
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    (Original post by ihatePE)
    can you leave it out like what you've done for all expressions like 4(2-x)(2-x) ?
    Yeah, you can leave it out for anything of the form \alpha(ax - b)(cx -d)(ex-f)\cdots (yx -z).
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    (Original post by Zacken)
    Yeah, you can leave it out for anything of the form \alpha(ax - b)(cx -d)(ex-f)\cdots (yx -z).
    amazing ive had this kind of expression in my calc gcse exam and was stumped
 
 
 
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