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

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

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

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

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2, 2, - 1

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Solution

The correct option is C $- 1, \frac{1}{2}, 2$
$f (x) = 2 x^{2} - 3 x^{2} - 3 x + 2$
$f (- 1) = - 2 - 3 + 3 + 2$
$f (- 1) = 0$
As $f - 1 = 0$
$∴ (x + 1)$ is a factor of $f (x)$
$f (x) = (x + 1) (2 x^{2} - 5 x + 2$
$f (x) = (x + 1) ({2 x}^{2} - 4 x - - x + 2)$
$f (x) = (x + 1) (2 x - 1) (x - 2)$
$f (x) = 0$
$∴ x = - 1, \frac{1}{2}, 2.$

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Similar questions

2, \frac{2}{3} x - \frac{5}{3} x^{2} + \frac{5}{2} x^{3}, \frac{- 4}{3} + \frac{2 x^{2}}{3} - \frac{x}{2}, \frac{5 x^{3}}{3} + 3 x^{2} + 3 x + \frac{6}{5}

A = {1, 2, 3}

R_{1} = {(1, 1), (2, 2)}

R_{2} = {(1, 1), (2, 2) (3, 3), (1, 3)}

R_{3} = {(1, 1), (2, 2), (3, 3)}

1 \pm \sqrt{- 2}, 2 \pm \sqrt{- 3}

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

\sqrt{2} x^{2} - 3 x - 2 \sqrt{2} = 0

g (x) = {\begin{matrix} x^{2}, & - 1 \leq x \leq 2 x + 2, & 2 < x \leq 3 \end{matrix}

f (x) = {\begin{matrix} x + 4, & x \leq 1 2 x + 1, & 1 < x \leq 2 \end{matrix}

f (g (x)) = 0

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