TS EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma, \delta\) are the roots of the equation \(12 x^4-56 x^3+89 x^2-56 x+12=0\) such that \(\alpha \beta=\gamma \delta=1\) and \(\frac{\alpha+\beta}{\gamma+\delta}>1\), then \(\frac{\alpha+\beta}{\gamma+\delta}=\)
- A \(\frac{65}{6}\)
- B \(\frac{13}{2}\)
- C \(\frac{17}{15}\)
- D \(\frac{15}{13}\)
Answer & Solution
Correct Answer
(D) \(\frac{15}{13}\)
Step-by-step Solution
Detailed explanation
The given equation is a reciprocal equation. Divide by \(x^2\) and substitute \(y = x + \frac{1}{x}\) (so \(x^2 + \frac{1}{x^2} = y^2 - 2\)). \(12(x^2 + \frac{1}{x^2}) - 56(x + \frac{1}{x}) + 89 = 0\) \(12(y^2 - 2) - 56y + 89 = 0\) \(12y^2 - 24 - 56y + 89 = 0\)…
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