A) \[0,\text{ }1\]
B) \[-1,\text{ }1\]
C) \[0,-1\]
D) \[-1,\text{ }2\]
Correct Answer: C
Solution :
One root of the quadratic equation \[{{x}^{2}}+px+(1-p)=0\]is\[(1-p)\]. So,\[(1-p)\]will satisfies this equation. \[\therefore \] \[{{(1-p)}^{2}}+p(1-p)+(1-p)=0\] \[\Rightarrow \] \[(1-p)[1-p+p+1]=0\] \[\Rightarrow \] \[2(1-p)=0\] \[\Rightarrow \] \[p=1\] On putting the value of p in given equation, we get \[{{x}^{2}}+x=0\] \[\Rightarrow \] \[x(x+1)=0\] \[\Rightarrow \] \[x=0,-1\]
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