C1 maths Help

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?

Almost there, just do (-1+√7) - (-1 -√7) to get 2√7, which is the length AB

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?

that's what i was stuck on i knew you had to something with the co-ordinates why not(-1 -√7)-(-1+√7)?

that's what i was stuck on i knew you had to something with the co-ordinates why not(-1 -√7)-(-1+√7)?

Quick way to check is just plug it back in

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

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.

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

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