Let p, q in mathbb{Q}. If 2 - sqrt{3} is a ro | vedclass

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

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

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(B) Given that $p, q \in \mathbb{Q}$ and $2 - \sqrt{3}$ is a root of the quadratic equation $x^2 + px + q = 0$.Since the coefficients are rational,the 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 $= -p$.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,the product of roots $= q$.So,$q = 1$.Now,check the options:For option $(B)$,$p^2 - 4q - 12 = (-4)^2 - 4(1) - 12 = 16 - 4 - 12 = 0$.Thus,option $(B)$ is correct.

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

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