Parabola problem

[Apricity]

Hi all!

I'm going crazy figuring out how to solve this problem. Please help!

For what values of k is the expression k(x+1)(x-8) - (x+4) negative definite?

I have expanded and then found the discriminant, but I don't get the right answer. I also know that k has to be less than 0 because it corresponds to 'a' in ax^2+bx+c... Thanks a lot for your help peeps

 

[HallsofIvy]

Hi all!

I'm going crazy figuring out how to solve this problem. Please help!

For what values of k is the expression k(x+1)(x-8) - (x+4) negative definite?

I have expanded and then found the discriminant, but I don't get the right answer. I also know that k has to be less than 0 because it corresponds to 'a' in ax^2+bx+c... Thanks a lot for your help peeps

What did you get when you expanded? what answer did you get?

 

[Apricity]

What did you get when you expanded? what answer did you get?

This is what I did:

k(x+1) (x-8) - (x+4)
k(x^2-7x-8) - (x+4)
kx^2-7kx-8k-x-4
kx^2-(7k+1)x-(8k+4)

I then figured that for it to be negative definite a <0 (k<0)...Then I tried getting the discriminant <0

b^2 -4ac<0

a=k
b=-(7k+1)
c=-(8k+4)

(-7k-1)^2 - 4[k(-8k-4)]<0
49k^2+14k+1 -4[-8k^2-4k]<0
49k^2+14k+1+32k^2+16k<0
81k^2+30k+1<0

This is where I'm stuck... Please help. Thanks!!

 

[Ishuda]

This is what I did: ...
81k^2+30k+1<0

This is where I'm stuck... Please help. Thanks!!

Almost there! Where is 81 k² + 30 k + 1 < 0? This is a quadratic equation in k, so when is it negative? Hint: If it is ever negative, it is between the zeros of the quadratic equation because it goes to +∞ as k goes to \(\displaystyle \pm\)∞

 

[Apricity]

Almost there! Where is 81 k² + 30 k + 1 < 0? This is a quadratic equation in k.


... Of course! I just had to use the quadratic formula I got -1/3 <k< -1/27 ...Thank you very much for your help!