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Weird question

  1. Jul 29, 2012 #1
    [itex]6pq+r=\sqrt{r^2-4ps}, \ \ p\neq 0 \ \ (1)[/itex]

    Without squaring (1) or any of it's rearrangements, express p in terms of only q, r and s.


    I realised this is some sort of quadratic formula so I got [tex] p = \dfrac{-r+\sqrt{r^2-4ps}}{6q} [/tex]

    so for the quadratic ap^2 + bp + c,
    a = 3q, b = r, c = s,
    3qp^2 + rp + s

    i'm stuck from here.
     
  2. jcsd
  3. Jul 29, 2012 #2

    SammyS

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    Does the quadratic formula count as squaring equation (1) ? Just asking ...

    If it's OK to use the quadratic formula, then notice that in relation to the standard quadratic equation,
    [itex]ax^2+bx+c=0[/itex]​
    the quantities under the radical are the coefficients, a, b, and c.
    [itex]\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/itex]​

    Puttying this in a form similar to eq. (1) gives
    [itex]\displaystyle 2ax+b=\pm\sqrt{b^2-4ac}[/itex]​
     
  4. Jul 29, 2012 #3

    gabbagabbahey

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    errm... [itex]4ac \neq 4ps[/itex] so this isn't quite so nice
     
  5. Jul 30, 2012 #4
    It doesn't no.

    I somehow managin to get [tex] p = \dfrac{-(rq+s)}{3q^2} [/tex] which is incorrect

    could you explain?
     
    Last edited: Jul 30, 2012
  6. Jul 30, 2012 #5
    never mind I rearranged into [itex] 3q = \dfrac{-r+\sqrt{r^2-4ps}}{2p} [/itex] and got the right answer.

    thanks!
     
    Last edited: Jul 30, 2012
  7. Jul 30, 2012 #6

    eumyang

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    Not sure how you got the right answer, because what you wrote doesn't follow from the original. Also, I thought the question was to...
    Color me confused. :confused:
     
  8. Jul 30, 2012 #7

    SammyS

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    So, what did you get for p ?
     
  9. Jul 30, 2012 #8


    well the original equation is [itex] 6pq + r = \sqrt{r^2-4ps} [/itex] I rearranged to get [itex] 3q = \dfrac{-r+\sqrt{r^2-4ps}}{2p} [/itex] I then compared it with [itex] x = \dfrac{-b+\sqrt{b^2-4ac}}{2a} [/itex] giving x = 3q, b = r, c = s so: [itex] p(3q)^2 +r3q +s = 0 [/itex] rearranging to get [itex]p = \dfrac{-(3qr + s)}{9q^2} [/itex] which is correct (or so it says.)
     
  10. Jul 30, 2012 #9

    SammyS

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    I agree.
     
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