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# C1 maths Help watch

1. Thanks for the tag thefatone but looks like someone beat me ;D
2. (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?
3. (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
4. (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)?
5. (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
6. (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
7. (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
8. (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...
9. (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?
Attached Images

Attached Files
10. q1.docx (415.3 KB, 41 views)
11. (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
12. (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
13. (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
14. (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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