TS EAMCET · Maths · Quadratic Equation
If \(\alpha\) and \(\beta\) are the real roots of the equation \(\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}\) and \(\alpha>\beta\), then \(\sqrt{\alpha^2-11^4 \beta^2}=\)
- A 64
- B 36
- C 100
- D 6
Answer & Solution
Correct Answer
(D) 6
Step-by-step Solution
Detailed explanation
We have, \(\sqrt{\frac{5 x}{x-2}}+\sqrt{\frac{x-2}{5 x}}=\frac{29}{10}\) Let \(\quad \sqrt{\frac{5 x}{x-2}}=y\) Then, \(y+\frac{1}{y}=\frac{29}{10} \Rightarrow \frac{y^2+1}{y}=\frac{29}{10}\) \(\Rightarrow \quad 10 y^2+10=29 y\) \(\Rightarrow \quad 10 y^2-29 y+10=0\)…
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