anonymous
  • anonymous
2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 need to determine A, B, C
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
We do it the same way. Let me work for a second here...
anonymous
  • anonymous
\[2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 \implies\]\[(2A+2C + 3B) +(6C+2B)x+2Cx^2=12x^2 \implies\]\[2C=12 \implies C=6\]\[6C+2B=0 \implies B=-18\]\[2A+2C + 3B=0 \implies A=9\]

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anonymous
  • anonymous
The corresponding terms on the left and right sides of the equation have to be the same. So, we match the x^2 coefficients to get C=6, then the x coefficients (note there is none on the RHS, so the LHS has to be zero) to get B=-18, then follow the same pattern to get A=9.
anonymous
  • anonymous
hey, check A, i think it's 21.
anonymous
  • anonymous
really, i'm getting it now, thanks a lot.
anonymous
  • anonymous
I think it's nine.... 3*6-2*18=-18 implies that A has to be pretty close to nine.
anonymous
  • anonymous
you interchanged the value of A and B.
anonymous
  • anonymous
OOPs!! Hate it when that happens!! Bad substitution!!
anonymous
  • anonymous
i noticed as well. it's alright! :D
anonymous
  • anonymous
You seem to have it now.... even if my arithmetic is lacking. LOL Do math every day.
anonymous
  • anonymous
you're still one great teacher! :)
anonymous
  • anonymous
sure, i'll do a lot of math everyday. ammm, do you do some DE's?
anonymous
  • anonymous
A little. What do you have in mind?
anonymous
  • anonymous
i'll post it on a new one, so that i can give you another medal.

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