A) \[{{a}^{2}}c+a{{c}^{2}}+{{b}^{3}}-3abc=0\]
B) \[{{a}^{2}}{{c}^{2}}+a{{c}^{2}}+{{b}^{2}}-3abc=0\]
C) \[a{{c}^{2}}+ac+{{b}^{3}}-3abc=0\]
D) \[{{a}^{2}}c+a{{c}^{2}}-{{b}^{3}}-3abc=0\]
E) \[ac+{{b}^{3}}-3abc=0\]
Correct Answer: A
Solution :
Let one root of equation\[a{{x}^{2}}+bx+c=0\]is\[\alpha ,\]then another root will be\[{{\alpha }^{2}}\]. \[\therefore \] \[\alpha +{{\alpha }^{2}}=-\frac{b}{a}\] ???(i) and \[\alpha .{{\alpha }^{2}}=\frac{c}{a}\] \[\Rightarrow \] \[{{\alpha }^{3}}=\frac{c}{a}\] From Eq.(i) \[{{(\alpha +{{\alpha }^{2}})}^{3}}=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\Rightarrow \] \[{{\alpha }^{3}}+{{\alpha }^{6}}+3{{\alpha }^{3}}(\alpha +{{\alpha }^{2}})=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\Rightarrow \] \[\frac{c}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}+3\frac{c}{a}\left( -\frac{b}{a} \right)=-\frac{{{b}^{3}}}{{{a}^{3}}}\] \[\Rightarrow \] \[{{a}^{2}}c+a{{c}^{2}}-3abc+{{b}^{3}}=0\]
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