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kx^2-5x+1=0 find range of values for k for which the equation will have real roots?

  • 01-12-2009 9:17pm
    #1
    Ste234
    Registered Users, Registered Users 2 Posts: 65 ✭✭


    In the above equation can some one tell me how to get an answer of k<6.25
    because I only get 12.5 which I'm obviously on the right track with?


Comments

  • #2
    Delphi91
    Registered Users, Registered Users 2 Posts: 1,501 ✭✭✭Delphi91


    Use the usual formula for a quadratic equation x = -b +-....etc

    If you do that, the term under the square root sign is 25 - 4k. For the roots to be real, this term must not be negative, i.e. it must be > 0.

    Therefore:

    25 - 4k > 0

    = 25 > 4k

    =4k < 25

    = k < 25/4

    = k < 6.25


  • #3
    Ste234
    Registered Users, Registered Users 2 Posts: 65 ✭✭Ste234


    Thanks, just realised I was making the most stupid mistake, I said b^2-2ac > 0
    rather than b^2-4ac > 0 for some reason.


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