Question

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + \frac{a}{2}x + b = 0$ and $(\alpha-\beta)(\alpha-\gamma)$, $(\beta-\alpha)(\beta-\gamma)$, $(\gamma-\alpha)(\gamma-\beta)$ are the roots of the equation $(y+a)^3 + K(y+a)^2 + L = 0$, then $\frac{L}{K}= $

• A. $\frac{32b^2}{a}$
• B. $\frac{16a^2}{b}$
• C. $\frac{18b^2}{a}$
• D. $\frac{12a^2}{b}$

Hint:

For roots expressed as differences of original roots, use the derivative $P'(\alpha)$ to simplify calculations for transformed polynomials.

Answer 1

The Correct Option is C

Step 1: Express roots in terms of derivative.
For $P(x) = x^3 + \frac{a}{2}x + b$ with roots $\alpha,\beta,\gamma$: \[ (\alpha-\beta)(\alpha-\gamma) = P'(\alpha) = 3\alpha^2 + \frac{a}{2}. \]
Step 2: Transform roots for the second equation.
Let $y_i = (\text{expression above})$, then $z_i = y_i + a = 3\alpha^2 + \frac{3a}{2}$. The second equation is: \[ z^3 + K z^2 + L = 0. \]
Step 3: Use Vieta's formulas.
Sum of roots: $\sum z_i = 3\sum \alpha^2 + \frac{9a}{2} = -3a + \frac{9a}{2} = \frac{3a}{2} \implies K=-3a/2$.
Step 4: Find product of roots.
\[ \prod z_i = 3^3 \prod (\alpha^2 + a/2) = 27 \prod (-b/\alpha) = 27 b^2 = -L \implies L = -27b^2. \]
Step 5: Compute $\frac{L{K}$.}
\[ \frac{L}{K} = \frac{-27b^2}{-3a/2} = \frac{18 b^2}{a}. \]