If 2 + isqrt{3} is a root of the equation x^2 | vedclass

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(A) Since the coefficients $p$ and $q$ are real,the complex roots must occur in conjugate pairs.Given that $2 + i\sqrt{3}$ is a root,its conjugate $2

Vedclass · Accepted Answer

$(-4, 7)$

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(A) Since the coefficients $p$ and $q$ are real,the complex roots must occur in conjugate pairs.Given that $2 + i\sqrt{3}$ is a root,its conjugate $2 - i\sqrt{3}$ must also be a root of the equation.For a quadratic equation $x^2 + px + q = 0$,the sum of the roots is given by $-p$ and the product of the roots is given by $q$.Sum of roots: $(2 + i\sqrt{3}) + (2 - i\sqrt{3}) = 4$. Thus,$-p = 4$,which implies $p = -4$.Product of roots: $(2 + i\sqrt{3})(2 - i\sqrt{3}) = 2^2 - (i\sqrt{3})^2 = 4 - (-3) = 4 + 3 = 7$. Thus,$q = 7$.Therefore,$(p, q) = (-4, 7)$.

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

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