Let p, q in Q. If 2 - sqrt{3} is a root of th | vedclass

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(C) Given that $p, q \in Q$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.Since the coefficients are rational,irrational r

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$p^2 - 4q - 12 = 0$

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(C) Given that $p, q \in Q$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.Since the coefficients are rational,irrational roots must occur in conjugate pairs.Therefore,the other root is $2 + \sqrt{3}$.Sum of roots $= (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$.From the equation $x^2 + px + q = 0$,the sum of roots is $-p$.So,$-p = 4 \implies p = -4$.Product of roots $= (2 - \sqrt{3})(2 + \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.From the equation $x^2 + px + q = 0$,the product of roots is $q$.So,$q = 1$.Now,check the options by substituting $p = -4$ and $q = 1$:$p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$.Thus,$p^2 - 4q - 12 = 0$ is the correct relation.

Let $p, q \in Q$. If $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$,then:

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