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Math Help - Solving the discriminant

  1. #1
    Senior Member Mukilab's Avatar
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    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?

    Thank you for your help
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Re: Solving the discriminant

    Quote Originally Posted by Mukilab View Post
    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?

    Thank you for your help
    You are Almost Surely (sorry for the pun I couldn't resist) correct c=-k-1
    Last edited by TheEmptySet; September 27th 2012 at 01:36 PM. Reason: spelling amost LOL
    Thanks from Mukilab
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  3. #3
    Senior Member Mukilab's Avatar
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    Re: Solving the discriminant

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


    Thanks for the help!
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    Re: Solving the discriminant

    Quote Originally Posted by Mukilab View Post
    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. The value of k that you give would mean that the quadratic would have two complex solutions.
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  5. #5
    Senior Member Mukilab's Avatar
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    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
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  6. #6
    Behold, the power of SARDINES!
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    Re: Solving the discriminant

    Quote Originally Posted by Mukilab View Post
    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
    Thanks from Mukilab
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