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Q. For $ a\ne b, $ if the equation $ {{x}^{2}}+ax+b=0 $ and $ {{x}^{2}}+bx+a=0 $ have a common root, then the value of $ a+b $ equals to:

KEAMKEAM 2004

Solution:

Let $ \alpha $ 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 $