A) 2 and\[-2\]
B) 2 and \[-1\]
C) 1 and\[-2\]
D) 1 and 2
E) \[-1\]and 2
Correct Answer: C
Solution :
\[\because \]a and b are roots of equation \[{{x}^{2}}+ax+b=0,a\ne 0,b\ne 0\] \[\therefore \] \[{{a}^{2}}+a.a+b=0\] \[\Rightarrow \] \[2{{a}^{2}}+b=0\] \[\Rightarrow \] \[b=-2{{a}^{2}}\] and \[{{b}^{2}}+a.b+b=0\] \[\therefore \] \[4{{a}^{4}}-2{{a}^{3}}-2{{a}^{2}}=0\] [using (i)] \[2{{a}^{2}}(2{{a}^{2}}-a-1)=0\] \[\Rightarrow \] \[2{{a}^{2}}=0\text{ }or\text{ }2{{a}^{2}}-a-1=0\] \[a=0\]or \[2{{a}^{2}}-2a+a-1=0\] \[\Rightarrow \] \[a=0\text{ }or\text{ (}2a+1)(a-1)=0\] \[\Rightarrow \,\,\,a=0\,\,or\,a=-\frac{12}{2}\,or\,a=1\] \[\therefore \] \[a=0\Rightarrow b=0\] \[a=-\frac{1}{2}\Rightarrow b=\frac{1}{2}\] \[a=1\Rightarrow b=-2\] Hence, required value of a and b are 1 and\[-2\]respectively.
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