TS EAMCET · Maths · Quadratic Equation
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 \((y+a)^3+\mathrm{K}(y+a)^2+\mathrm{L}=0\), then \(\frac{\mathrm{L}}{\mathrm{K}}=\)
- A \(\frac{32 b^2}{a}\)
- B \(\frac{16 a^2}{b}\)
- C \(\frac{18 b^2}{a}\)
- D \(\frac{12 a^2}{b}\)
Answer & Solution
Correct Answer
(C) \(\frac{18 b^2}{a}\)
Step-by-step Solution
Detailed explanation
\(\sum \alpha = 0 \implies \sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 0^2 - 2(\frac{a}{2}) = -a\) \(Z_k = 3\alpha_k^2 + \frac{3a}{2}\) \(-\mathrm{K} = \sum Z_k = 3\sum \alpha_k^2 + 3(\frac{3a}{2}) = 3(-a) + \frac{9a}{2} = \frac{3a}{2}\) \(\mathrm{K} = -\frac{3a}{2}\)…
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