A) Positive
B) Negative
C) Real
D) Imaginary
Correct Answer: C
Solution :
Given equation\[(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0\] can be re-written as \[3{{x}^{2}}-2(a+b+c)x+(ab+bc+ca)=0\] \[\Delta =4\left\{ {{(a+b+c)}^{2}}-3(ab+bc+ca) \right\}(\because {{b}^{2}}-4ac=\Delta )\] \[=4({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)\] \[=2\left\{ {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}} \right\}\ge 0\] Hence both roots are always real.
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