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AP EAMCET · Maths · Quadratic Equation

If \(x_1, x_3\) are the roots of \(A x^2-4 x+1=0\) and \(x_2, x_4\) are the roots of \(B x^2-6 x+1=0\) such that \(x_1, x_2, x_3, x_4\) are in harmonic progression, then \(\frac{B+A}{B-A}=\)

  1. A \(\frac{11}{5}\)
  2. B \(\frac{-11}{5}\)
  3. C \(\frac{5}{11}\)
  4. D \(\frac{-5}{11}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{11}{5}\)

Step-by-step Solution

Detailed explanation

Let \(y_i = \frac{1}{x_i}\). Since \(x_1, x_2, x_3, x_4\) are in HP, \(y_1, y_2, y_3, y_4\) are in AP. For \(A x^2-4 x+1=0\), roots are \(x_1, x_3\). Reciprocal roots \(y_1, y_3\) satisfy \(y^2-4y+A=0\). \(y_1+y_3=4\) \(y_1 y_3=A\) For \(B x^2-6 x+1=0\), roots are \(x_2, x_4\).…