• # question_answer Let $p,q\in R.$If $2-\sqrt{3}$ is a root of the quadratic equation, ${{x}^{2}}+px+q=0,$ then : [JEE Main 9-4-2019 Morning] A) ${{q}^{2}}+4p+14=0$   B) ${{p}^{2}}4q12=0$C) ${{q}^{2}}4p16=0$          D) ${{p}^{2}}4q+12=0$

In given question $p,q\in R.$ If we take other root as any real number $\alpha ,$then quadratic equation will be ${{x}^{2}}-(\alpha +2-\sqrt{3})x+\alpha .\left( 2-\sqrt{3} \right)=0$ Now, we can have none or any of the options can be correct depending upon$'\alpha '$ Instead of $p,q\in R$it should be $p,q\in Q$then other root will be $2+\sqrt{3}$ $\Rightarrow p=-\left( 2+\sqrt{3}-2-\sqrt{3} \right)=-4$ and$q=\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=1$ $\Rightarrow {{p}^{2}}-4q-12={{(-4)}^{2}}-4-12$                         $=16-16=0$ Option [b] is correct