Let p ,   q ∈   Q   . If 2 - 3 is a root of the quadratic equation x 2 + p x + q = 0 ,…

In given equation $x^2+px+q=0, p, q\in Q.$

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 $-\frac{b}{a}$ and $\frac{c}{a}.$

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

$\Rightarrow p=-4$

And, the product of roots $\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=q$

$\Rightarrow q=2^2-{\left(\sqrt{3}\right)}^2=1$

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

$=16-16=0.$

Thus, the answer is, $p^2-4q-12=0.$