lilMissMindset 2 years ago 2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 need to determine A, B, C

1. lilMissMindset

@AnimalAin this

2. AnimalAin

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

3. AnimalAin

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

4. AnimalAin

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.

5. lilMissMindset

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

6. lilMissMindset

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

7. AnimalAin

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

8. lilMissMindset

you interchanged the value of A and B.

9. AnimalAin

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

10. lilMissMindset

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

11. AnimalAin

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

12. lilMissMindset

you're still one great teacher! :)

13. lilMissMindset

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

14. AnimalAin

A little. What do you have in mind?

15. lilMissMindset

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