Here's the question you clicked on:
lilMissMindset
2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 need to determine A, B, C
We do it the same way. Let me work for a second here...
\[2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 \implies\]\[(2A+2C + 3B) +(6C+2B)x+2Cx^2=12x^2 \implies\]\[2C=12 \implies C=6\]\[6C+2B=0 \implies B=-18\]\[2A+2C + 3B=0 \implies A=9\]
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.
hey, check A, i think it's 21.
really, i'm getting it now, thanks a lot.
I think it's nine.... 3*6-2*18=-18 implies that A has to be pretty close to nine.
you interchanged the value of A and B.
OOPs!! Hate it when that happens!! Bad substitution!!
i noticed as well. it's alright! :D
You seem to have it now.... even if my arithmetic is lacking. LOL Do math every day.
you're still one great teacher! :)
sure, i'll do a lot of math everyday. ammm, do you do some DE's?
A little. What do you have in mind?
i'll post it on a new one, so that i can give you another medal.