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 $q = ( 2 + \sqrt{3} ) (2 - \sqrt{3}) = 1$
$\Rightarrow \, p^2 - 4q - 12 = (-4)^2 - 4 - 12$
$= 16 - 16 = 0$