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Topic: rearranging a quadratic equaton  (Read 3931 times)

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Offline rcole23

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rearranging a quadratic equaton
« on: August 09, 2009, 09:42:30 PM »


I please ask if someone can check up on my negative and positive signs?  I think I have made mistakes there.

0.04 (1.6 - x) (2.4 - x) = 4 x2

0.04 + 3.84 - 1.6x + 2.4x + x2 = 4x2

(.064 - .04x)(2.4 - x)

foil

.1536 - .064x - .096x + .04x2 = 4x2

.1536 - .064x - .096x + .04x2 - 4x2 = 0

.1536 - 0.16x - 3.96x2 = 0

Offline AWK

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Re: rearranging a quadratic equaton
« Reply #1 on: August 10, 2009, 03:17:01 AM »


I please ask if someone can check up on my negative and positive signs?  I think I have made mistakes there.

0.04 (1.6 - x) (2.4 - x) = 4 x2

0.04 + 3.84 - 1.6x + 2.4x + x2 = 4x2

(.064 - .04x)(2.4 - x)

foil

.1536 - .064x - .096x + .04x2 = 4x2

.1536 - .064x - .096x + .04x2 - 4x2 = 0

.1536 - 0.16x - 3.96x2 = 0
AWK

Offline rcole23

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Re: rearranging a quadratic equaton
« Reply #2 on: August 10, 2009, 01:49:23 PM »
hmm how about this then?

0.04 (1.6 - x) (2.4 - x) = 4 x2

0.04 (3.84 - 1.6x - 2.4x + x2) = 4x2

.1536 - .064x - .096x + .04x2 = 4x2

.1536 - .064x - .096x + .04x2 - 4x2 = 0

.1563 - .016x - 3.96x2 = 0


Offline zxt

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Re: rearranging a quadratic equaton
« Reply #3 on: August 10, 2009, 01:53:18 PM »
right

Offline rcole23

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Re: rearranging a quadratic equaton
« Reply #4 on: August 10, 2009, 01:56:16 PM »
Actually what about ?

0.04 (1.6 - x) (2.4 - x) = 4 x2

0.04 (3.84 + 1.6x + 2.4x + x2) = 4x2

.1536 + .064x + .096x + .04x2 = 4x2

.1536 + .064x + .096x + .04x2 - 4x2 = 0

.1563 + .016x - 3.96x2 = 0

Final answer 3.96x2 + .016x - .1536 = 0

Perhaps the - is not incurring that the number is negative simply meaning that its - (+ #)  What do you guys think?  The other answer and the signs remain static.. 

I posted on another forum and the above answer is the one they supplied with the signs switched from my own answer.  I did the math in minus a positive multiplication in mind and I received the answer they did.  Not sure which one is correct now.

I am a little confused about the positive and negatives.   I have followed the rules I know, any one able to confirm or provide advice?  P.S. Thanks ZXT

Offline rcole23

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Re: rearranging a quadratic equaton
« Reply #5 on: August 10, 2009, 02:05:50 PM »
Somone even posted this answer

3.96 x2 + 0.16 x - 0.1536 = 0

which is also different than mine ;( 

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