Question

Been working on the following question for a lot of pages, but each answer just doesn't full fill the equation. I think im doing something in correctly See reply for question. ( Solving a quadratic equation )

Answer 1

If \[\frac{ 3c^2 + 3xc + x^2 }{ 3x^2 + 3yc + y^2 } = \frac{yV _{1}}{yV _{2}}\]
find the positive value of c when

x=4, y=6, V1 = 120, V2 = 314

Show work.

Answer 2

So far, what I have been doing is the following. \[\frac{3c^2 + 3 * 4c + 4^2}{3c^2 + 3*6c + 6^2} = \frac{6*120}{4*315}\]\[\frac{3c^2 + 12c + 16}{3c^2 + 18c + 36} = (\frac{720}{1260}) = \frac{4}{7} \]\[7(3c^2 + 12c + 16) = 4(3c^2 + 18c + 36) \] and I belive either there on, or right there I am making a critical mistake.

Answer 3

Correction, V2 is 315. *

Answer 4

You just need to make sure that\[3*c^2+18*c+36\] is not zero

Answer 5

Not sure if I understood that correctly, considering Finding the positive number would assume that c > 0. c would not be 0.

The next steps I've tried go on \[3*7c^2 + 12*7c + 16 *7 = 3*4c^2 + 18*4c + 36*4 \]\[21c^2 + 84c + 112 = 12c^2 + 72c + 144 \]\[21c^2 - 12c^2 + 84c - 72c +112 - 144 = 0\]\[9c^2 +12c - 32 = 0 \] and thats where i usually get stuck, as I am not sure how would I continue (tried completing the square, but difficulties arise.

Answer 6

Figured it up on my own, feeling stupid, but for those curious the steps after that are the following

\[3c*3c + 3 * -4c + 3 * 8 c + -4*8 = 0 \]\[(3c+8) (3c-4) = 0\] thus
(3c+8) = 0
3c = -8
\[c = - \frac{8}{3}\]or

3c-4 = 0
3c = 4
\[c = \frac{4}{3}\]

since c > 0,
answer is \[c = \frac{4}{3}\]