question_answer

If \[\operatorname{a}+b+c=0\], then the solution of the equation

A) 0                                 

B)   \[\pm \frac{3}{2}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\]

C)   \[0,\,\,\pm \,\sqrt{\frac{3}{2}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\] 

D)   \[0,\,\,\pm \sqrt{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\]

Correct Answer: C

Solution :

\[\Rightarrow \,\,\,\,x=\sum a=0\] \[or\,\,1\{(b\,-x)(c-x)-{{a}^{2}}\}-c\{c-x-a\}+b\{a-b+x\}=0\] (by expanding the determinant.) \[\operatorname{or}\,\,{{x}^{2}}-({{a}^{2}}+{{b}^{2}}+{{c}^{2}})+(ab-bc+ca)=0\] or \[{{\operatorname{x}}^{2}}-\sum {{a}^{2}}\,\,+\,\,\sum ab\,\,=\,\,0\] or \[{{\operatorname{x}}^{2}}-\,\,\left( \sum {{a}^{2}} \right)\,-\frac{1}{2}\,\,\left( \sum {{a}^{2}} \right)\,\,\,=\,\,0\] \[[\because \,\,a+b+c=0\,\,\Rightarrow \,\,{{(a+b+c)}^{2}}=\,\,0\] \[\Rightarrow \,\,\,\sum {{a}^{2}}+2\sum ab=0\,\,\Rightarrow \,\,\sum ab=-\frac{1}{2}\sum {{a}^{2}}]\] \[x=\pm \sqrt{\frac{3}{2}\sum {{a}^{2}}}\] \[\therefore \] the solution is \[\therefore \,\,\,the\,\,solution\,\,is\,\,x=0\,\,or\,\,\pm \,\,\sqrt{\frac{3}{2}\sum {{a}^{2}}}\]