C1 short question - Has me baffled

f(x) = x^2+4kx+(3+11k) where k is a constant

1. (I have no problem with this question but i'll just mention it)
Express f(x) in the form (x+p)^2 + q, where p andq are constants to be found in terms of k.
Ans: (x+2k)^2 - 4k^2 +(3+11k)

2. (the one I have a problem with )

Given that the equation f(x) = 0 has no real roots
find the set of possible values ofk.

easy right.

b^2 - 4ac <0
therefore
4k^2 - 4(3+11k) <0
when I open up the brackets I get 4k^2 -44k - 12 <0
However, the mark scheme says I must ave 4k^2 -11k- 3 and k must be -1/4, 3 :s-smilie:

What did I do wrong?

Reply 1

username537390

The

b^2

part of

b^2 - 4ac

b = 4k

, so what is

(4k)^2

Reply 2

Jonario

Original post by dnumberwang

The

b^2

part of

b^2 - 4ac

b = 4k

, so what is

(4k)^2

16k^2 0.o ?

Reply 3

MostCompetitive

Original post by Jonario

16k^2 0.o ?

correct.

Reply 4

Jonario

And I then divide by 4 -.-

Thanks guys

Why is 4K b? Why isn’t 11?

Reply 7

ManLike007

Original post by Duckified

Why is 4K b? Why isn’t 11?

f(x)=x^{2}+4kx+(3k+11)

is in the form of

ax^{2}+bx+c

a=1, b=4k, c=3k+11

Reply 8

Duckified

Original post by ManLike007

f(x)=x^{2}+4kx+(3k+11)

is in the form of

ax^{2}+bx+c

a=1, b=4k, c=3k+11

Ah right okay, thanks

Reply 9

julietlima3

Original post by Duckified

Why is 4K b? Why isn’t 11?

anything that isn't a coefficient of x (ie. a constant) is the 'c' part of ax^2+bx+c

because we know k is a constant, then 3k+11 must also be a constant