Theory of Equations 1 Question 2
2. Let $p, q \in \mathbf{R}$. If $2-\sqrt{3}$ is a root of the quadratic equation, $x^{2}+p x+q=0$, then
(2019 Main, 9 April I)
(a) $q^{2}-4 p-16=0$
(b) $p^{2}-4 q-12=0$
(c) $p^{2}-4 q+12=0$
(d) $q^{2}+4 p+14=0$
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Solution:
- Given quadratic equation is $x^{2}+p x+q=0$, where $p, q \in \mathbf{R}$ having one root $2-\sqrt{3}$, then other root is $2+\sqrt{3}$ (conjugate of $2-\sqrt{3}$ ) [ $\because$ irrational roots of a quadratic equation always occurs in pairs]
So, sum of roots $=-p=4 \Rightarrow p=-4$
and product of roots $=q=4-3 \Rightarrow q=1$
Now, from options $p^{2}-4 q-12=16-4-12=0$
