If one root of $24 x^{3} - 14 x^{2} - 63 x + 45 = 0$ is double the other then the roots are

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$- 1, \frac{1}{2}, 2$

$2, 2, - 1$

$\frac{3}{4}, \frac{3}{2}, - \frac{5}{3}$

$- \frac{3}{2}, - \frac{3}{4}, - \frac{1}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t α, 2 α, β a r e r o o t s 24 x^{3} - 14 x^{2} - 63 x + 45 = 0 \\ S_{1} = α + 2 α + β = \frac{14}{24} \\ \Rightarrow 3 α + β = \frac{14}{24} = \frac{7}{12} - - (1) \\ S_{2} = 2 α^{2} + 2 α β = \frac{- 63}{24} = \frac{- 21}{8} - - (2) \\ S_{3} = 2 α^{2} β = \frac{- 45}{24} = \frac{- 15}{8} - - (3) \\ f r o m (1) & (2) \\ \Rightarrow 8 α^{2} - 2 α - 3 = 0 \\ \Rightarrow α = \frac{3}{4} o r \frac{- 1}{2} \\ f r o m (1) \\ β = \frac{- 5}{3} o r \frac{25}{12} \\ α = \frac{- 1}{2}, β = \frac{25}{12} n o t s a t i s f y (3) \\ ∴ α = \frac{3}{4}, β = \frac{- 5}{3} \\ ∴ T h e r o o t s a r e \frac{3}{4}, \frac{3}{2}, \frac{- 5}{3} \end{array}$