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If two of the roots of $2 x^{3} - 3 x^{2} - 3 x + 2 = 0$ are differ by 3 then roots are

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

2, 2, - 1

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

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

answer is C.

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Detailed Solution

2 x^{3} - 3 x^{2} - 3 x + 2 = 0

Question Image

$(x - 2) (2 x^{2} + x - 1)$

Take $2 x^{2} + x - 1 = 0$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$\frac{- 1 \pm \sqrt{1 - 4 (2) (- 1)}}{4} = \frac{- 1 \pm 3}{4} = - 1, \frac{1}{2}$

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

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