A) 1
B) \[-1\]
C) \[-6\]
D) 6
Correct Answer: B
Solution :
Put \[{{x}^{2}}+x=y\] So that, the equation becomes \[(y-2)(y-3)=12\] \[\Rightarrow \] \[{{y}^{2}}-5y-6=0\] \[\Rightarrow \] \[(y-6)(y+1)=0\] \[\therefore \] \[y=6,-1\] When\[y=6,\]then we get \[{{x}^{2}}+x-6=0\] \[\Rightarrow \] \[(x+3)(x-2)=0\] \[\therefore \] \[x=-3,2\] When\[y=-1,\]then we get \[{{x}^{2}}+x+1=0\] \[\Rightarrow \] \[x=\omega ,{{\omega }^{2}}\] and their sum is\[\omega +{{\omega }^{2}},\]i.e., \[(-1)\] \[(\because 1+\omega +{{\omega }^{2}}=0)\]
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