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C1 Edexcel Jan2010 Question! Please help!

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Blizz_rb93:
Hi everyone, sorry for my inactivity lately. Been studying for my mocks and externals :(

So I have this C1 mock exam tomorrow and I've been solving pastpapers all day, on the January 2010 exam I came across this question that I didn't know how to solve, I checked the mark scheme but it was confusing, can anyone help me?
Here's the question

Question 10 :

f(x) = x2 + 4kx + (3 + 11k), where k is a constant

a) express f(x) in the form (x+p)2 + q, where p and q are constants to be found in terms of K.

Given that the equation f(x) = 0 has no real roots,

b) find the set of possible values of k.

Given that k=1,
c) Sketch the graph of y=f(x), showing the coordinates of any point at which the graph crosses a coordinate axis..

Please anyone help me, any replies will be VERY APPRECIATED!!

Thank you all in advance!!

albiol18:
do u want the mark scheme?

Angel.:
Okay, I'm going to try this :) :

a) x^2 + 4kx = (x + 2k)^2 - 4k^2   therefore     x^2 + 4kx + (3 + 11k) = (x + 2k)^2 - 4k^2 + 3 + 11k
therefore p = 2k and q = - 4k^2 + 3 + 11k

b) Since f(x) = 0 has no real roots, b^2 - 4ac < 0
a = 1, b = 4k and c = (3 + 11k)
therefore (4k)^2 - 4(1)(3 + 11k) < 0
therefore 16k^2 - 44k - 12 < 0
then divide by 4, so therefore 4k^2 - 11k - 3 < 0
therefore factorise so (4k + 1)(k - 3) < 0
therefore k = -1/4 or k = 3
therefore -1/4 < k < 3

c) k = 1 therefore using (x + 2k)^2 - 4k^2 + 3 + 11k, you know that y = (x + 2)^2 - 4 + 3 + 11 = (x + 2)^2 + 10
when x = 0, y = 14 therefore the curve cuts the y-axis at (0,14)
the curve will not cut the x-axis because f(x) = 0 has no real roots
the lowest point of the curve will be (-2,10)

I hope this helps :)

Blizz_rb93:

--- Quote from: Angel. on April 21, 2010, 05:27:41 pm ---thank you for your reply man, but how did you get this part? i dont understand :/

a) x^2 + 4kx = (x + 2k)^2 - 4k^2  

--- End quote ---

Angel.:
No problem :D

I got that by completing the square.
In the case above, you have x^2 + 4kx and you know that it will end up looking something like (x + ...)^2 + constant. When you complete the square, you have to halve the b value (which in this case is 4k) so that that can go in the bracket. Then you get (x + 2k)^2 although that will give you an extra 4k^2 when you expand the bracket. Therefore you need to minus 4k^2 from this and you end up getting (x + 2k)^2 - 4k^2.

I'm sorry if that's an awful explanation, but I can't really explain it through words properly. If my reasoning makes no sense, you can Google 'completing the square', although I'm sure that you probably know how to do it ;)

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