# Solving the discriminant

• Sep 27th 2012, 12:10 PM
Mukilab
Solving the discriminant
If f(x)=2x^2-3x-(k+1)

Do I take c in the discriminant (b^2-4ac) as (k+1) or (-k-1)?

Surely I should take it as -k-1?

• Sep 27th 2012, 12:15 PM
TheEmptySet
Re: Solving the discriminant
Quote:

Originally Posted by Mukilab
If f(x)=2x^2-3x-(k+1)

Do I take c in the discriminant (b^2-4ac) as (k+1) or (-k-1)?

Surely I should take it as -k-1?

You are Almost Surely (sorry for the pun I couldn't resist) correct $c=-k-1$
• Sep 27th 2012, 12:32 PM
Mukilab
Re: Solving the discriminant
So the answer would be k<17/8? (from 9-4(2)(-k-1) )

Thanks for the help!
• Sep 27th 2012, 12:35 PM
TheEmptySet
Re: Solving the discriminant
Quote:

Originally Posted by Mukilab
So the answer would be k<17/8? (from 9-4(2)(-k-1) )

Thanks for the help!

Since I don't know the entire question that is hard to answer. :p The value of $k$ that you give would mean that the quadratic would have two complex solutions.
• Sep 27th 2012, 12:39 PM
Mukilab
Re: Solving the discriminant
Sorry! The question is:
The equation 2x^2-3x-(k+1)=0 where k is a constant, has no real roots. Find the set of possible values of k.

Therefore the discriminant must be <0 for f(x) to have no real roots. Therefore discriminant=(-3)^2-4(-k-1)(2)=17+8k

Therefore 17+8k<0
Therefore 8k<-17
Therefore k<-17/8
• Sep 27th 2012, 12:44 PM
TheEmptySet
Re: Solving the discriminant
Quote:

Originally Posted by Mukilab
Sorry! The question is:
The equation 2x^2-3x-(k+1)=0 where k is a constant, has no real roots. Find the set of possible values of k.

Therefore the discriminant must be <0 for f(x) to have no real roots. Therefore discriminant=(-3)^2-4(-k-1)(2)=17+8k

Therefore 17+8k<0
Therefore 8k<-17
Therefore k<-17/8

Yes you are correct