A) positive
B) negative
C) real
D) complex
Correct Answer: C
Solution :
We have, the equation \[(x-b)(x-c)+(x-c)(x-a)\] \[+(x-a)(x-b)=0\] \[\Rightarrow \] \[{{x}^{2}}-cx-bx+bc+{{x}^{2}}-ax-cx+ac\] \[+{{x}^{2}}-bx-ax+ab=0\] \[\Rightarrow \] \[3{{x}^{2}}-2(a+b+c)x+ab+bc+ca=0\] \[\Delta ={{B}^{2}}-4AC\] \[=4{{(a+b+c)}^{2}}-4(3)(ab+bc+ca)\] \[=4[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac\] \[-3ab-3bc-3ca]\] \[=4[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-(ab+bc+ca)]>0\] \[\because \] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}>ab+bc+ca\] \[\therefore \] Roots are real.
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