TS EAMCET · Maths · Quadratic Equation
The roots of the equation \(x^3-14 x^2+56 x-64=0\) are in
- A arithmetic-geometric progression
- B harmonic progression
- C arithmetic progression
- D geometric progression
Answer & Solution
Correct Answer
(D) geometric progression
Step-by-step Solution
Detailed explanation
\(x^3-14 x^2+56 x-64=0\) \(\begin{aligned} & \alpha+\beta+\gamma=14 \\ & \alpha \beta+\beta \gamma+\gamma \alpha=56 \\ & \alpha \beta \gamma=64 \end{aligned}\) By trial, \(x=2\) is a root of equation…
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