If $\alpha, \beta$ are the roots of the equation $ax^{2} + bx + c = 0$, then $log (a - bx + cx^{2})$ is equal to
- $\log a + (\alpha + \beta) x + \dfrac {\alpha^{2} + \beta^{2}}{2} x^{2} + \dfrac {\alpha^{3} + \beta^{3}}{3}$
- $\log a + (\alpha + \beta) x - \left (\dfrac {\alpha^{2} + \beta^{2}}{2}\right ) \cdot x^{2} + \left (\dfrac {\alpha^{3} + \beta^{3}}{3}\right )x^{3}$
- $\log a - (\alpha + \beta) x - \left (\dfrac {\alpha^{2} + \beta^{2}}{2}\right ) x^{2} - \left (\dfrac {\alpha^{3} + \beta^{3}}{3}\right )x^{3}$
-
None of the above
Using the Taylor expansion for log(1-u) = -u - u^2/2 - u^3/3..., the expression log(a(1 - (b/a)x + (c/a)x^2)) = log(a) + log(1 - (b/a)x + (c/a)x^2). Given roots alpha, beta, (b/a) = alpha+beta and (c/a) = alpha*beta. The expansion matches option B.
Rewrite the quadratic inside the logarithm as a(1 - (b/a)x + (c/a)x^2). Using the fact that the sum of roots is b/a and the product is c/a, this becomes a(1 - alpha*x)(1 - beta*x). Apply the logarithm to get log(a) + log(1 - alpha*x) + log(1 - beta*x). Expand each term using the standard logarithmic series log(1 - z) = -z - z^2/2 - z^3/3, which yields log(a) + (alpha + beta)x - ((alpha^2 + beta^2)/2)x^2 + ((alpha^3 + beta^3)/3)x^3.