WBJEE · Maths · Quadratic Equation
Let \(x_{1}, x_{2}\) be the roots of \(x^{2}-3 x+a=0\) and \(x_{3}, x_{4}\) be the roots of \(x^{2}-12 x+b=0\) If \(x_{1} < x_{2} < x_{3} < x_{4}\) and \(x_{1}, x_{2}, x_{3}, x_{4}\) are in GP. then ab equals
- A \(\frac{24}{5}\)
- B 64
- C 16
- D 8
Answer & Solution
Correct Answer
(B) 64
Step-by-step Solution
Detailed explanation
We have, \(x_{1}, x_{2}\) be the roots of equation \(x^{2}-3 x+a=0\) \(\therefore \quad x_{1}+x_{2}=3\) and \(x_{1} x_{2}=a\) Also, \(x_{3}, x_{4}\) be the roots of equation \(x^{2}-12 x+b=0\) \(\therefore x_{3}+x_{4}=12\) and \(x_{3} x_{4}=b\) Again,…
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