A) 3
B) 4
C) 2
D) 1
Correct Answer: A
Solution :
Let \[\alpha \]and 3 are the roots of the equation \[{{x}^{2}}+ax+3=0\] \[\therefore \] \[3\alpha =3\Rightarrow \alpha =1\] and \[3+\alpha =-a\Rightarrow a=-4\] Again, let \[\beta \]and 3\[\beta \] are the roots of the equation \[{{x}^{2}}+ax+b=0\]. \[\therefore \] \[\beta +3\beta =-a\]\[\Rightarrow 4\beta =4\] \[[\because a=-4]\] \[\Rightarrow \] \[\beta =1\] and \[\beta .3\beta =b\Rightarrow 3\times 1=b\Rightarrow b=3\]You need to login to perform this action.
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