# [SOLVED] Please check my rearranging

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• April 11th 2010, 07:40 PM
Lucy1
[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?
• April 11th 2010, 08:10 PM
maddas
I get 50.
• April 11th 2010, 08:28 PM
amul28
ur just doing a calculation mistake while substituting the value of p...
u get answer as 50 check again
• April 11th 2010, 08:41 PM
Lucy1
yeah thanks dudes. Been overloading my mind with calcs so couldnt see what i was doing!

Now i get 50.