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Question

If 2+i3 is a root of the equation x2+px+q=0, where p and q are real, then (p, q) =


A

(-4,-7)

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B

(4,7)

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C

(4,7)

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D

(-4,7)

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Solution

The correct option is D

(-4,7)


Since 2+i3 is a root, therefore 2i3 will be other root. Now sum of the roots = 4 = -p and product of roots = 7 = q. Hence (p, q) = (-4, 7).


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