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    Well I know the basics but im stuck on a problem I've had a look on the mark scheme but that just confused me like twice as much
    I know you use B^2-4ac=0
    No clue what to do after that :s
    Thx

    xD forgot to put in question
    Theres the link (not sure how to put it as a pic)
    http://s54.radikal.ru/i146/1011/05/a86e9c0ccc2c.jpg
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    Well, you know that  b^2 -4ac = 0 so that just means  (2p)^2 - 4(3p+4) = 0 \implies 4p^2 - 12p - 16 = 0 which can be solved quadratically, (you know p > 0 remember) part (b) should follow easily then.
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    (Original post by Scott3142)
    Well, you know that  b^2 -4ac = 0 so that just means  (2p)^2 - 4(3p+4) = 0 \implies 4p^2 - 12p - 16 = 0 which can be solved quadratically, (you know p > 0 remember) part (b) should follow easily then.
    but theres a X next to the 2p so when i tried that i included the Xs why would you take the X out :s
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    It's simply because the quadratic equation  x = \frac{-b\pm \sqrt{b^2 -4ac}}{2a} is given as the solution to  ax^2 + bx + c = 0 (ie, it's found by rearranging to get x as the subject) so in this case, the  2p is just your "b" so you don't need the x.
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    (Original post by Scott3142)
    It's simply because the quadratic equation  x = \frac{-b\pm \sqrt{b^2 -4ac}}{2a} is given as the solution to  ax^2 + bx + c = 0 (ie, it's found by rearranging to get x as the subject) so in this case, the  2p is just your "b" so you don't need the x.
    o.o thanks I think i get it
 
 
 
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