The sum of all non-integer roots of the equation $x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0$ is

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Solution

The correct option is B 3
Given: $x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0$

Equation can be written as: $\begin{matrix} x^{4} (x - 1) - 5 x^{3} (x - 1) + 6 x^{2} (x - 1) + x (x - 1) - 2 (x - 1) = 0 \Rightarrow (x - 1) (x^{4} - 5 x^{3} + 6 x^{2} + x - 2) = 0 \Rightarrow (x - 1) (x^{4} - 2 x^{3} - 3 x^{3} + 6 x^{2} + x - 2) = 0 \Rightarrow (x - 1) (x^{3} (x - 2) - 3 x^{2} (x - 2) + 1 (x - 2)) = 0 \Rightarrow (x - 1) (x - 2) (x^{3} - 3 x^{2} + 1) = 0 \end{matrix}$
Equation $(x^{3} - 3 x^{2} + 1 = 0)$ has all the non-integer roots.
Thus, sum of the all the non-integer roots of the above equation
$= - \frac{coeff. of x^{2}}{coeff. of x^{3}} = - \frac{- 3}{1} = 3$

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Similar questions

x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0

x^{5} - 6 x^{4} + 11 x^{3} - 5 x^{2} - 3 x + 2 = 0

x^{4} - 3 x^{3} - 2 x^{2} + 3 x + 1 = 0

U = {x | x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}

A = {x | x^{2} - 5 x + 6 = 0}

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

(A \cap B)^{'}

U = {x | x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}

A = {x | x^{2} - 5 x + 6 = 0}

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

(A \cap B)^{'}

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