A) 0, 1
B) \[-1,1\]
C) 0,-1
D) \[-1,2\]
Correct Answer: C
Solution :
Since\[(1-p)\]is the root of quadratic equation \[{{x}^{2}}+px+(1-p)=0\] ...(i) So,\[(1-p)\]satisfied the above equation. \[\therefore \] \[{{(1-p)}^{2}}+p(1-p)+(1-p)=0\] \[(1-p)[1-p+p+1]=0\] \[\Rightarrow \] \[(1-p)(2)=0\] \[\Rightarrow \] \[p=1\] On putting this value of p in Eq. (i) \[{{x}^{2}}+x=0\] \[\Rightarrow \] \[x(x+1)=0\] \[\Rightarrow \] \[x=0,-1\]You need to login to perform this action.
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