A) \[\frac{2}{3},1\]
B) \[\frac{2}{3},6\]
C) \[0,1\]
D) \[0,2\]
E) \[\frac{2}{3},0\]
Correct Answer: B
Solution :
\[{{x}^{2}}-(2+m)x+({{m}^{2}}-4m+4)=0\] Since,\[x\]has equal roots, then \[{{B}^{2}}-4AC=0\] \[{{(m+2)}^{2}}-4({{m}^{2}}-4m+4)=0\] \[\Rightarrow \] \[({{m}^{2}}+4+4m)-(4{{m}^{2}}-16m+16)=0\] \[\Rightarrow \]\[-3{{m}^{2}}+20m-12=0\] \[\Rightarrow \]\[3{{m}^{2}}-20m+12=0\] \[\Rightarrow \]\[3{{m}^{2}}-18m-2m+12=0\] \[\Rightarrow \]\[3m(m-6)-2(m-6)=0\] \[\Rightarrow \] \[(m-6)(3m-2)=0\] \[\Rightarrow \] \[m=\frac{2}{3},6\]
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