2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2

need to determine A, B, C

Question

2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2

need to determine A, B, C

anonymous · Answer

@AnimalAin  this

anonymous · Answer

We do it the same way.  Let me work for a second here...

anonymous · Answer

$$2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 \implies$$$$(2A+2C + 3B) +(6C+2B)x+2Cx^2=12x^2 \implies$$$$2C=12 \implies C=6$$$$6C+2B=0 \implies B=-18$$$$2A+2C + 3B=0 \implies A=9$$

anonymous · Answer

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 · Answer

hey, check A, i think it's 21.

anonymous · Answer

really, i'm getting it now, thanks a lot.

anonymous · Answer

I think it's nine.... 3*6-2*18=-18 implies that A has to be pretty close to nine.

anonymous · Answer

you interchanged the value of A and B.

anonymous · Answer

OOPs!!  Hate it when that happens!! Bad substitution!!

anonymous · Answer

i noticed as well. it's alright! :D

anonymous · Answer

You seem to have it now.... even if my arithmetic is lacking.  LOL  Do math every day.

anonymous · Answer

you're still one great teacher! :)

anonymous · Answer

sure, i'll do a lot of math everyday. ammm, do you do some DE's?

anonymous · Answer

A little.  What do you have in mind?

anonymous · Answer

i'll post it on a new one, so that i can give you another medal.