The repeated root of the equation $4 x^{3} - 12 x^{2} - 15 x - 4 = 0$ is

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$\frac{5}{2}$

$- \frac{1}{3}$

$- \frac{1}{2}$

$\frac{1}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} L e t α, α, β a r e r o o t s o f 4 x^{3} - 12 x^{2} - 15 x - 4 = 0 \\ ∴ 2 α + β = \frac{12}{4} \Rightarrow 2 α + β = 3 - - (1) \\ α^{2} + 2 α β = \frac{- 15}{4} - - (2) \\ ∴ α^{2} β = \frac{4}{4} \Rightarrow α^{2} β = 1 - - (3) \\ f r o m (1) & (2) \\ 2 α + \frac{1}{α^{2}} = 3 \\ \Rightarrow 2 α^{3} - 3 α^{2} + 1 = 0 \\ \Rightarrow (α - 1) (2 α^{2} - α - 1) = 0 \\ ∴ α - 1 = 0, 2 α^{2} - α - 1 = 0 \\ \Rightarrow 2 α^{2} - 2 α + α - 1 = 0 \\ \Rightarrow α = 1, \frac{- 1}{2} \\ W h e n α = 1 t h e n β = 1 n o t s a t i s i f c i e (2) \\ W h w n α = \frac{- 1}{2} t h e n β = 4 s a t i s i f i c e (2) \\ ∴ α = \frac{- 1}{2} i s r e p a t e d r o o t \end{array}$