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    Thanks for the tag thefatone but looks like someone beat me ;D
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    (Original post by Student403)
    Thanks for the tag thefatone but looks like someone beat me ;D
    so far for question 9b i've integrated for part a and it tells me the graph goes through the origin so constant=0
    y=6x-2x²-x³

    y=x(6-2x-x²)

    then i used the quadratic formula and got 2±√28/-2 = -1±√7 is that right?
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    (Original post by thefatone)
    so far for question 9b i've integrated for part a and it tells me the graph goes through the origin so constant=0
    y=6x-2x²-x³

    y=x(6-2x-x²)

    then i used the quadratic formula and got 2±√28/-2 = -1±√7 is that right?
    Almost there, just do (-1+√7) - (-1 -√7) to get 2√7, which is the length AB
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    (Original post by Dapperblook22)
    Almost there, just do (-1+√7) - (-1 -√7) to get 2√7, which is the length AB
    that's what i was stuck on i knew you had to something with the co-ordinates why not(-1 -√7)-(-1+√7)?
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    (Original post by thefatone)
    so far for question 9b i've integrated for part a and it tells me the graph goes through the origin so constant=0
    y=6x-2x²-x³

    y=x(6-2x-x²)

    then i used the quadratic formula and got 2±√28/-2 = -1±√7 is that right?
    Quick way to check is just plug it back in
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    (Original post by thefatone)
    that's what i was stuck on i knew you had to something with the co-ordinates why not(-1 -√7)-(-1+√7)?
    Because you can't have a negative length, this would give -2√7, which does not make sense unless you consider vectors or displacement
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    (Original post by thefatone)
    that's what i was stuck on i knew you had to something with the co-ordinates why not(-1 -√7)-(-1+√7)?
    Would give you -2rt7 which is basically the same thing. You take the modulus as it is a length and can't be negative
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    (Original post by Student403)
    Quick way to check is just plug it back in

    no way am i doing that with someting looking like that xD

    (Original post by Student403)
    Would give you -2rt7 which is basically the same thing. You take the modulus as it is a length and can't be negative

    ah right distance isn't negative...
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    (Original post by Dapperblook22)
    Attachment 514435

    Please find attached my working for part 9b, I am certain this one is right this time though because it is actually in the mark scheme and I checked it this time :P

    EDIT: Oops, some of it was truncated. The answer at the end should be 2root7 as the length AB is found by subtracting the B value from the A value.
    Attachment 514439514441 i reached till here, don't i have to square root AB after subtracting it?
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  1. File Type: docx q1.docx (415.3 KB, 33 views)
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    (Original post by YesterdaysDreams)
    Attachment 514439514441 i reached till here, don't i have to square root AB after subtracting it?
    You won't need to use the formula for distance between two points, just treat it as lengths along a ruler for example
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    (Original post by Dapperblook22)
    You won't need to use the formula for distance between two points, just treat it as lengths along a ruler for example
    okay,thanks
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    (Original post by YesterdaysDreams)
    okay,thanks
    Also I just noticed a slight error in your distance formula, you could have used it, but the actual formula is



    Your formula forgot the squared signs
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    (Original post by Dapperblook22)
    Also I just noticed a slight error in your distance formula, you could have used it, but the actual formula is



    Your formula forgot the squared signs
    yeah I just noticed that,
 
 
 
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