If the roots of $x^{4} - 10 x^{3} + 37 x^{2} - 60 x + 36 = 0$ are $α, α, β, β, (α < β)$ then $2 α + 3 β - 2 αβ =$

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–4

–1

answer is B.

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Detailed Solution

$\begin{array}{l} α, α, β, β a r e r o o t s o f x^{4} - 10 x^{3} + 37 x^{2} - 60 x + 36 = 0 \\ ∴ α + α + β + β = 10 \Rightarrow 2 (α + β) = 10 \Rightarrow α + β = 5 - - (1) \\ ∴ α . α . β . β = 36 \\ \Rightarrow {(α β)}^{2} = 36 \Rightarrow α β = 6 \\ α - β = \pm \sqrt{{(α + β)}^{2} - 4 α β} = \pm \sqrt{25 - 24} = \pm 1 - - (2) \\ (1) + (2) \\ 2 α = 6 \Rightarrow α = 3 o r 2 α = 4 \Rightarrow α = 2 \\ (1) - (2) \\ 2 β = 4 \Rightarrow β = 2 o r 2 β = 6 \Rightarrow β = 3 \\ ∵ α < β \\ ∴ α = 2, β = 3 \\ 2 α + 3 β - 2 α β = 2 (2) + 3 (3) - 2 (2) (3) \Rightarrow 13 - 12 = 1 \end{array}$