Question

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

• A. $p^2-4q+12=0$
• B. $q^2-4p-16=0$
• C. $p^2-4q-12=0$
• D. $q^2+4p+14=0$

Answer 1

The correct option is C $p^2-4q-12=0$

Let $\alpha$ be the second root of the quadratic equation, then equation will be:
$x^2-(\alpha +2-\sqrt{3})x+\alpha (2-\sqrt{3})=0$
But we can see that none of the option satisfies this equation.
Hence, other root will be $2+\sqrt{3}$.
$\Rightarrow p=-(2+\sqrt{3}+2-\sqrt{3})=-4$
and $q=(2+\sqrt{3})(2-\sqrt{3})=1$
By putting values of $p$ and $q$ in all the options, we can see that option (c) holds true for $p$ and $q$.