AP EAMCET · Maths · Quadratic Equation
If the roots of the equation \(x^5-40 x^4-P x^3-R x-S=0\) are in geometric progression and the sum of the reciprocals of the roots is 10 , then \(|\mathrm{S}|=\)
- A \(8\)
- B \(16\)
- C \(32\)
- D \(64\)
Answer & Solution
Correct Answer
(C) \(32\)
Step-by-step Solution
Detailed explanation
\(\Sigma x_i = 40\) \(\Sigma \frac{1}{x_i} = 10\) \(x_1 x_2 x_3 x_4 x_5 = S\) Let roots be \(k/q^2, k/q, k, kq, kq^2\). \(\Sigma x_i = k(q^{-2}+q^{-1}+1+q+q^2) = 40\) \(\Sigma \frac{1}{x_i} = \frac{1}{k}(q^2+q+1+q^{-1}+q^{-2}) = 10\) \(k^2 = 40/10 = 4\) \(k = \pm 2\)…
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