If \(\frac{a}{c}\) = \(\frac{c}{d}\) = k, find the value of \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)
A. 3k
2B. 2k - k
2C. \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
D. k
2 Correct Answer: C
Explanation
\(\frac{a}{c}\) = \(\frac{c}{d}\) = k
∴ \(\frac{a}{b}\) = bk
\(\frac{c}{d}\) = k
∴ c = dk
= \(\frac{3a^2 - ac + c^2}{3b^2 - bd + d^2}\)
= \(\frac{3(bk)^2 - (bk)(dk) + dk^2}{3b^2 - bd + a^2}\)
= \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
k = \(\frac{3b^2k^2 - bk^2d + dk^2}{3b^2 - bd + d^2}\)
