A) \[-1\]
B) 0
C) 1
D) 2
E) \[-2\]
Correct Answer: A
Solution :
Let a be the common root for both the equations\[{{x}^{2}}+ax+b=0\]and\[{{x}^{2}}+bx+a=0,\] then \[{{\alpha }^{2}}+a\alpha +b=0\] ...(i) and \[{{\alpha }^{2}}+b\alpha +a=0\] ...(ii) \[\Rightarrow \] \[\frac{{{\alpha }^{2}}}{\left| \begin{matrix} a & b \\ b & a \\ \end{matrix} \right|}=\frac{\alpha }{\left| \begin{matrix} b & 1 \\ a & 1 \\ \end{matrix} \right|}=\frac{1}{\left| \begin{matrix} 1 & a \\ 1 & b \\ \end{matrix} \right|}\] \[\Rightarrow \] \[\frac{{{\alpha }^{2}}}{({{a}^{2}}-{{b}^{2}})}=\frac{\alpha }{b-a}=\frac{1}{b-a}\] \[\therefore \] \[{{\alpha }^{2}}=-(a+b)\]and \[\alpha =1\] Hence, \[a+b=-1\]
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