The roots of the equation $x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$ are

1, 1, 1, 1

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2, 2, 2, 2

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3, 1, 3, 1

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1, 2, 1, 2

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Solution

The correct option is A
1, 1, 1, 1

Let $f (x) = x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0$

⇒ $(x - 1) (x^{3} - 3 x^{2} + 3 x - 1 = 0)$

⇒ $(x - 1) (x - 1)^{3} = 0$ ⇒ x = 1, 1, 1, 1

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Similar questions

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

\sqrt{- 1}

- 1

x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1 = 0

Solve the equation $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5 = 0$ , given one root is $\sqrt{(- 1)}$ . Find the value of other three roots.

If one root of the equation $x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 5$ is $\sqrt{- 1}$ . Find the sum of the squares of other three roots.

α, β, γ, σ

x^{4} + 4 x^{3} - 6 x^{2} + 7 x - 9 = 0

(1 - α^{2}) (1 - β^{2}) (1 - γ^{2}) (1 - σ^{2})

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