Question

Let $p,q\in R$. If $2-\sqrt{3}$ is a root of the quadratic equation, $x^2+px+q=0$, then :

• A. $q^2+4p+14=0$
• B. $p^2-4p-12=0$
• C. $q^2-4p-14=0$
• D. $p^2-4p+12=0$

Answer 1

Answer: (B)

In given question $p,q\in R$. If we take other root
as any real number $\alpha$,
then quadratic equation will be
$x^2-(\alpha +2-\sqrt{3})x+\alpha .(2-\sqrt{3})=0$
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 $2+\sqrt{3}$
$\Rightarrow p=-(2+|\sqrt{3}-2-\sqrt{3})=-4$
and $q=(2+\sqrt{3})(2-\sqrt{3})=1$
$\Rightarrow p^2-4q-12=(-4{)}^2-4-12$
$=16-16=0$