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TS EAMCET · Maths · Quadratic Equation

If \(\alpha, \beta\) are the roots of \(x^2+a x+2=0\) and \(\frac{1}{\alpha}, \frac{1}{\beta}\) are the roots of \(x^2-b x+c=0\), then \(\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)=\)

  1. A \(\frac{9}{4}\left(9-a^2\right)\)
  2. B \(\frac{9}{4}\left(9+a^2\right)\)
  3. C \(\frac{9}{4}\left(9-b^2\right)\)
  4. D \(\frac{9}{4}\left(9+b^2\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{9}{4}\left(9-a^2\right)\)

Step-by-step Solution

Detailed explanation

\(x^2+a x+2=0\) \(\begin{aligned} & \alpha+\beta=-a, \alpha \beta=2 \\ & x^2-b x+c=0\end{aligned}\)…