A) positive
B) real
C) imaginary
D) negative
Correct Answer: B
Solution :
The given equation is \[(x-b)(x-c)+(x-a)(x-b)+(x-a)\] \[+(x-c)=0\] \[\Rightarrow \]\[3{{x}^{2}}-2(a+b+c)x+(ab+bc+ca)=0\] \[\therefore \]Disc \[=4{{(a+b+c)}^{2}}-12(ab+bc+ca)\] \[=4\{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab+bc-ca\}\] \[=2\{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}\}\] Which is always positive. Hence both the roots of given equation are real. Note: If the value of discriminent is\[\ge 0\] or \[<0,\] then the roots of an equation are real or imaginary respectively.
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