JEE Main 2019 — Mathematics Algebra

In given equation ${x}^{2}+px+q=0,p,q\in Q.$x2+px+q=0,p,q∈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 ${x}^{2}+px+q=0$x2+px+q=0 is $2-\sqrt{3},$2−3​, then, the other root will be $2+\sqrt{3}.$2+3​.

We know that the sum and product of the roots of a quadratic equation $a{x}^{2}+bx+c=0$ax2+bx+c=0 are respectively $-\frac{b}{a}$−ab​ and $\frac{c}{a}.$ac​.

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

$$\Rightarrow p=-4$$⇒p=−4

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

$$\Rightarrow q={2}^{2}-{(\sqrt{3})}^{2}=1$$⇒q=22−(3​)2=1

Thus, we have ${p}^{2}-4q-12={(-4)}^{2}-4\times 1-12$p2−4q−12=(−4)2−4×1−12

$$=16-16=0.$$=16−16=0.

Thus, the answer is, ${p}^{2}-4q-12=0.$p2−4q−12=0.