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Topic: rearranging into quadratic form  (Read 7237 times)

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Offline mandy9008

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rearranging into quadratic form
« on: February 20, 2009, 05:13:03 PM »
i am trying to find out how to find the to the following problem:


when rearranging the equation:

0.820=x/(1.00-x)(2.00-x)

rearranging into quadratic form   ax^2+bx+c=0

when i rearranged it i got
a=.820
b=-1
c=1.64

but the answer for b=-3.46, how is this possible.
Please explain it to me!

Offline Pirate

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Re: rearranging into quadratic form
« Reply #1 on: February 20, 2009, 05:27:12 PM »
i am trying to find out how to find the to the following problem:


when rearranging the equation:

0.820=x/(1.00-x)(2.00-x)

rearranging into quadratic form   ax^2+bx+c=0

when i rearranged it i got
a=.820
b=-1
c=1.64

but the answer for b=-3.46, how is this possible.
Please explain it to me!

0,820 = x / (1-2x)(2-x)
0,820(1-2x)(2-x)=x
0,820(x^2-3x+2)=x
0,820x^2-2,46x+2,82-x=0
0,820x^2-3,46x+2,82=0

then just use quadratic equation to find out x

Offline Ak

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Re: rearranging into quadratic form
« Reply #2 on: February 20, 2009, 05:28:39 PM »
solving the bottom of the right side gives x^2 - 3x +2

so now it says .820 = x/(x^2 - 3X +2)

bring the bottom over and get

                    0.820x^2 - 2.46X + 1.64 = X
bring the x over and u get
                  0.820x^2 - 3.46X + 1.64 = 0

Edit: Pirate beat me 2 it

Offline mandy9008

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Re: rearranging into quadratic form
« Reply #3 on: February 20, 2009, 05:31:36 PM »
okay. thanks.
the problem was that i didn't FOIL the (1.00-x)(2.00-x)
THANKS SO MUCH!!

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