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

x2+(2+i3+2−i3)x+(2+i3)(2−i3)=0p=−(α+β), q=αβp=−4q=22−(i3)2=4+3=7(p,q)=(−4,7)

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

x2+(2+i3+2−i3)x+(2+i3)(2−i3)=0p=−(α+β), q=αβp=−4q=22−(i3)2=4+3=7(p,q)=(−4,7)

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If $2 + i \sqrt{3}$ is a root of the equation $x^{2} + p x + q = 0$ , then $(p, q) =$

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$(- 4, 7)$

$(4, 7)$

$(- 4, - 7)$

$(4, - 7)$

answer is C.

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Detailed Solution

$x^{2} + (2 + i \sqrt{3} + 2 - i \sqrt{3}) x + (2 + i \sqrt{3}) (2 - i \sqrt{3}) = 0$

$p = - (α + β)$ , $q = α β$

$p = - 4$

$q = 2^{2} - {(i \sqrt{3})}^{2}$

$= 4 + 3$

$= 7$

$(p, q) = (- 4, 7)$

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