Let p, q ∈ Q . If 2 - 3 is a root of the quadratic equation x ^ 2 +...

In given equation ${x}^{2} + px + q = 0, p, q \in Q .</p><p>And, we know that$ the irrational roots of a quadratic equation exist in conjugate pair, if the coefficients are rational.

Thus, if one root of the equation{x}^{2} + px + q = 0$is$2 - \sqrt{3},$then, the other root will be$2 + \sqrt{3} .

We know that the sum and product of the roots of a quadratic equation $a{x}^{2} + bx + c = 0$ are respectively $- \dfrac{b}{a} and$\dfrac{c}{a} .

Therefore, the sum of roots$2 + \sqrt{3} + 2 - \sqrt{3} = - p$

$⇒ p = - 4</p><p>And, the$product of roots \left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right) = q$</p><p>$⇒ q = {2}^{2} - {\left(\sqrt{3}\right)}^{2} = 1

Thus, we have ${p}^{2} - 4q - 12 = {\left(- 4\right)}^{2} - 4 \times 1 - 12$

$= 16 - 16 = 0 .</p><p>Thus, the answer is,${p}^{2} - 4q - 12 = 0 .