A) equal
B) imaginary
C) real and distinct
D) rational and equal
Correct Answer: C
Solution :
Given, \[\left( x-a)(x-a-1 \right)+\left( x-a-1 \right)\left( x-a-2 \right)\] \[+\left( x-a \right)\left( x-a-2 \right)=0\] Let \[x-a=t,\]then \[t\left( t-1 \right)+\left( t-1 \right)\left( t-2 \right)+t\left( t-2 \right)=0\] \[\Rightarrow \] \[{{t}^{2}}-t+{{t}^{2}}-3t+2+{{t}^{2}}-2t=0\] \[\Rightarrow \] \[3{{t}^{2}}-6t+2=0\] \[\Rightarrow \] \[t=\frac{6\pm \sqrt{36-24}}{2(3)}=\frac{6\pm 2\sqrt{3}}{2(3)}\] \[\Rightarrow \] \[x-a=\frac{3\pm \sqrt{3}}{3}\] \[\Rightarrow \] \[x=a+\frac{3\pm \sqrt{3}}{3}\] Hence, \[x\] is real and distinct.
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