A) \[{{q}^{2}}+4p+14=0\]
B) \[{{p}^{2}}4q12=0\]
C) \[{{q}^{2}}4p16=0\]
D) \[{{p}^{2}}4q+12=0\]
Correct Answer: B
Solution :
In given question \[p,q\in R.\] If we take other root as any real number \[\alpha ,\]then quadratic equation will be \[{{x}^{2}}-(\alpha +2-\sqrt{3})x+\alpha .\left( 2-\sqrt{3} \right)=0\] 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 \[2+\sqrt{3}\] \[\Rightarrow p=-\left( 2+\sqrt{3}-2-\sqrt{3} \right)=-4\] and\[q=\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=1\] \[\Rightarrow {{p}^{2}}-4q-12={{(-4)}^{2}}-4-12\] \[=16-16=0\] Option [b] is correct
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