The equation whose roots are increased by $1$ than those of $x^{4} + 1 = 0$ , is:

x^{4} + 1 = 0

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

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

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x^{4} - 1 = 0

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Solution

The correct option is A $x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 2 = 0$
In $x^{4} + 1 = 0$
Replacing $x \to x - 1$ , we get
${(x - 1)}^{4} + 1 = 0$
$\Rightarrow (x - 1)^{2} (x - 1)^{2} + 1 = 0$
$\Rightarrow x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 2 = 0$

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Similar questions

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

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.

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

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