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# Solve this following

Question:

Let $p, q \in R$. If $2-\sqrt{3}$ is a root of the quadratic

equation, $x^{2}+p x+q=0$, then :

1. $\mathrm{q}^{2}+4 \mathrm{p}+14=0$

2. $\mathrm{p}^{2}-4 \mathrm{q}-12=0$

3. $q^{2}-4 p-16=0$

4. $p^{2}-4 q+12=0$

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