A) \[\frac{4}{49}\]
B) \[\frac{49}{4}\]
C) \[4\]
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
Given, 4 is a root of the equation \[{{x}^{2}}+px+12=0\] \[\therefore \] \[16+4p+12=0\] \[\Rightarrow \] \[4p=-28\] \[\Rightarrow \] \[p=-7\] And equation\[{{x}^{2}}+px+q=0\]have equal roots, let roots are\[\alpha ,\alpha \]. \[\therefore \] \[\alpha +\alpha =-\frac{p}{1}=-p\] \[\Rightarrow \] \[2\alpha =-p\] \[\Rightarrow \] \[\alpha =-\frac{p}{2}=\frac{7}{2}\] \[[\because p=-7]\] and \[\alpha .\alpha =q\] \[\Rightarrow \] \[{{\left( \frac{7}{2} \right)}^{2}}=q\] \[\Rightarrow \] \[q=\frac{49}{4}\]
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