The equation whose roots are1, 2±3i is

x-2+3ix-2-3ix-1=(x2−4x+13)(x−1)=0 ⇒x3−5x2+17x−13=0

The equation whose roots are1, 2±3i is

x-2+3ix-2-3ix-1=(x2−4x+13)(x−1)=0 ⇒x3−5x2+17x−13=0

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The equation whose roots are $1, 2 \pm 3 i$ is

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

$x^{3} - x^{2} - 3 x + 3 = 0$

$x^{3} - 4 x^{2} - 8 x + 8 = 0$

$x^{3} - 5 x^{2} + 17 x - 13 = 0$

answer is D.

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Detailed Solution

$(x - (2 + 3 i)) (x - (2 - 3 i)) (x - 1) = (x^{2} - 4 x + 13) (x - 1) = 0$

$\Rightarrow x^{3} - 5 x^{2} + 17 x - 13 = 0$

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The equation whose roots are1, 2±3i is

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The equation whose roots are $1, 2 \pm 3 i$ is