A) \[4\]
B) \[12\]
C) \[3\]
D) \[\frac{29}{4}\]
E) \[\frac{49}{4}\]
Correct Answer: E
Solution :
Let 4 and a be the roots of the equation \[{{x}^{2}}+px+12=0\]. \[\therefore \] \[4a=12\Rightarrow \alpha =3\] And \[4+3=-p\Rightarrow p=-7\] \[\therefore \]Equation\[{{x}^{2}}+px+q=0\]will reduce to \[{{x}^{2}}-7x+q=0\] Let this equation have P, P as its roots. \[\therefore \] \[2\beta =7\Rightarrow \beta =\frac{7}{2}\]and\[{{\beta }^{2}}=q\] \[\Rightarrow \] \[q={{\left( \frac{7}{2} \right)}^{2}}=\frac{49}{4}\]
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