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