Solve the equation:
$2 x^{4} + x^{3} - 6 x^{2} + x + 2 = 0$ .

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Solution

Given equation, $2 x^{4} + x^{3} - 6 x^{2} + x + 2 = 0$
$⟹ 2 (x^{2} + x^{- 2}) + (x + x^{- 1}) - 6 = 0 [dividing by x^{2}]$
Substitute $x + x^{- 1} = y$ in the above equation
$⟹ 2 (y^{2} - 2) + y - 6 = 0 ⟹ 2 y^{2} + y - 6 = 0 ⟹ (2 y + 5) (y - 2) = 0$
$⟹ x + x^{- 1} = \frac{- 5}{2} and x + x^{- 1} = 2$

Solving the 2 quadratic equations, we have $1, 1, - 2, \frac{- 1}{2}$ as the roots of the given equation

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