URGENT!!!!!! math paper 1 cie 2008 may june question 8

[superduper2009]

can sum1 please do and explain question 8 in may/june 2008 paper 1
i'll b really gr8full ...
thanks alot
superduper2009

[astarmathsandphysics]

1st thing when i get up

[cashem'up]

can sum1 please do and explain question 8 in may/june 2008 paper 1
i'll b really gr8full ...
thanks alot
superduper2009

step 1  find fg(x)
2. f(9/2-x)
3  4(9/2-x) - 2k
 so u found the expansion now simplify it
 36/2-x -2k
ok now equate it to x as Q says

36/2-x -2k = x
2k= 36/2-x -x
2k = 36 - (2x -x²) / 2-x
4k -2kx = 36 - 2x + x²
x² - 2x +2kx + 36-4k=0
x² +(-2+2k)x + (36- 4k )

now we have a quadratic equation........ now as we know when b² - 4ac< 0 then no roots
b² - 4ac=0 then u get two equal roots
b² - 4ac>0 and two distinct roots

so  here
(2k-k)² - (4x1x(36-4k) = 0
u will then get a quadratic solve it and u will get the values

k=5 and -7

Thanks