Question

Given that y = 4kx^2 - 6x + (k-4), find the set of values for which y is negative for all real values of x.

What do I do???? :P

Answer 1

set of values of k ?

Answer 2

^Yup :)

Answer 3

first off \(k<0\) since otherwise you have a parabola that opens up and it will be positive eventually

Answer 4

then make sure the discriminant is negative, so it will have no real zeros

Answer 5

then the parabola will live entirely below the \(x\) axis

Answer 6

i .e. make sure \(b^2-4ac<0\) with
\[a=4k,b=-6,c=k-4\]

Answer 7

you should be good from there right?

Answer 8

hey that will give complex roots

Answer 9

exactly

Answer 10

meaning you have a parabola that lies entirely below the \(x\) axis i.e. \(y<0\) for all \(x\)

Answer 11

OH THAT MAKES SENSE. ILY SATELLITE. REALLY.

Answer 12

thanx
you still have a bunch more work to do
you have to solve
\[36-16k(k-4)<0\] for \(k<0\)

Answer 13

my ans is k<-0.5, k>2/9 ???

Answer 14

good work, but don't forget that \(k<0\) too

Answer 15

so only \(k<-.5\)

Answer 16

huh??? why that??

Answer 17

your leading coefficient is \(4k\) and if the leading coefficient is positive this will open up
you want to make sure it opens down, so you need \(k<0\)

Answer 18

anyhoo, forget this, and just tell me if it said y is positive for all real values of x, then it would be similar, as there would be no real roots, right???

Answer 19

hypothetical, only :P

Answer 20

and of the coefficient of x^2 was positive. oops.

Answer 21

exactly and in fact it if it says only positive you have already solved that.
it would be \(k>\frac{2}{9}\)

Answer 22

hahahahaha, THANK YOU. I AM FOREVER IN YOUR DEBT. No, siriusly. I have this imp math tutorial thing :P

Answer 23

good luck!

Answer 24

thanks :)