If one root of the equation {x^2} + px + q = | vedclass

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(A) Since the coefficients of the quadratic equation ${x^2} + px + q = 0$ are assumed to be rational,if one root is $2 + \sqrt{3}$,the other root must

Vedclass · Accepted Answer

$-4, 1$

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(A) Since the coefficients of the quadratic equation ${x^2} + px + q = 0$ are assumed to be rational,if one root is $2 + \sqrt{3}$,the other root must be its conjugate,$2 - \sqrt{3}$.Sum of the roots $= (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.From the equation,the sum of the roots is $-p$.Therefore,$-p = 4$,which implies $p = -4$.Product of the roots $= (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.From the equation,the product of the roots is $q$.Therefore,$q = 1$.Thus,the values are $(p, q) = (-4, 1)$.

If one root of the equation ${x^2} + px + q = 0$ is $2 + \sqrt{3}$,then the values of $p$ and $q$ are:

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