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Math Help - [SOLVED] Please check my rearranging

  1. #1
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    [SOLVED] Please check my rearranging

    ok so i start of with this
     50 = \frac{2p^2}{(0.4-p)(0.1-p)}
    so rearranging a bit

     50= \frac{2p^2}{p^2-0.5p+0.04}

     2p^2 = 50(p^2-0.5p+0.04)

     2p^2= 50p^2-25p+2

     48p^2-25p+2=0

    ok so i put in the qudratic forumula and i come out with
    x= 0.42212
    x=0.09871

    my problem is that when i put this value into the original equation i get 100 not 50. What am i doing wrong?
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  2. #2
    Senior Member
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    I get 50.
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  3. #3
    Member
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    ur just doing a calculation mistake while substituting the value of p...
    u get answer as 50 check again
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  4. #4
    Newbie
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    yeah thanks dudes. Been overloading my mind with calcs so couldnt see what i was doing!

    Now i get 50.
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