Maths Ocr 2016 C1

Just did the June 2007 FP1 paper (grade boundary for an A was 54) and was far far easier

how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?

Question 1?! Please someone help

Expand the brackets on the right to get x^2 +2x +11= 2kx-k

Move to one side to get = 0, so x^2 +(2- 2k)x+11+k= 0

Sub the values into b^2-4ac > 0
a=1, b=2-2k, c=11+k

Then solve to get k^2-3k-10 = 0, factorise to get (k-5)(k+2)>0,

Draw a quadratic graph that intercepts x axis at k=5 + k=-2, so k<-2 and k>5

Hope this helps

how did people find k in the question which was something like x^2 + 2x +11 = k(2x-1)?

That was one of the questions I didn't think was too bad but that I most likely got wrong anyway. Basically put everything on one side so that would be:
X^2 +2X -2kx +11 +K=0 (think I went wrong here put +2KX and -K aka simple adding and subtraction error :/)
You then you the discriminant and I believe the question specified that there would be 2 real roots so b^2-4ac>0.
Substitute in the values of a, b and c:
(2+2K)^2 -4x1x(11+K)>0
4K^2 +8K +4 -44-4K>0
4K^2+4K-44>0
Now what I did here was divide by 4 which should be correct.
K^2 +K -11>0
So as this is a positive K squared graph, I said there would be 2 separate inequalities as the graph would dip below the x-axis therefore 2 distinct areas where K is greater than 0.
I also used completing the square to solve this which may not have been a good idea...anyway I got:
(k+0.5)^2 -0.25-11>0
K>-0.5+11.25^0.5
K<-0.5-11.25^0.5
(^0.5 slightly award to write compared to square root but I don't know how to express that in the forum's text).
So yeah don't think this is right and I'm pretty sure I made a stupid error in putting 10 instead of 11 or something so I would have lost the one of the method marks I may have got for that last step....

How many marks do you think I will get for the last question I got 18root27/12 which is equivalent to 27 but will I lose marks for not having it as a whole number? Thanks