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 one root is $x_1 = 2 + i\sqrt{3}$,the other roo

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 one root is $x_1 = 2 + i\sqrt{3}$,the other root must be $x_2 = 2 - i\sqrt{3}$. Using the properties of quadratic equations,the sum of the roots is $x_1 + x_2 = (2 + i\sqrt{3}) + (2 - i\sqrt{3}) = 4$. From the equation $x^2 + px + q = 0$,the sum of the roots is $-p$. Therefore,$-p = 4$,which implies $p = -4$. The product of the roots is $x_1 	imes x_2 = (2 + i\sqrt{3})(2 - i\sqrt{3}) = 2^2 - (i\sqrt{3})^2 = 4 - (-3) = 7$. From the equation $x^2 + px + q = 0$,the product of the roots is $q$. Therefore,$q = 7$. Thus,$(p, q) = (-4, 7)$.

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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