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Quadratic Question

  1. Jan 8, 2008 #1
    Quadratic Question [Urgent]

    The equation [tex]x^{2} + 2px + (3p + 4) = 0[/tex], where p is a postitve constant, has equal roots

    Q) Find the value of p

    Well if it has equal roots b² - 4ac = 0
    so

    (2p)² - 4(1)(3p + 4) = 0

    4p² - 12p - 16 = 0

    4(p² - 3p - 4) = 0

    4((p+1)(p-4)) = 0

    so the anwser is p = 16

    BUT the question booklet says p = 4

    Which is right???
    Thx :)
     
  2. jcsd
  3. Jan 8, 2008 #2
    How did you get 16? Also there are two values of p that make the statement true.
     
  4. Jan 8, 2008 #3
    yeh i know the there are 2 valeus but p is a positive constant, as it says in the question

    erm i got p = 4
    but then u get the 4 otside the brackets so i multiply p by 4 giving me p = 16 or do you not count the four outside the brackets?
     
  5. Jan 8, 2008 #4

    Vid

    User Avatar

    Well, you can either divide both sides by 4 first to get p = 4 or if you wanted to keep the 4.
    4(p-4) =0
    4p-16 = 0
    p = 4
     
  6. Jan 8, 2008 #5
    that was a stupid mistake on my part
    sorry
    yeh it was soooo simple

    thanks
    :)
     
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