If the quadratic equations $x^2+ax+b=0\text{ and }{x}^2+bx+a=0 (a\ne b)$ have a common root, the numerical value of $a + b$ is . . . 

Given equations are 

   $x^2+ax+b=0 ...\left(i\right)$

and $x^2+bx+a=0 ...\left(ii\right)$

Subtracting Eqs. (i) and (ii), we get 

   $(a-b)x=(a-b)$

$\Rightarrow x=1$

Put $x =1,$ in Eq. (i), we get 

$\begin{array}{r}a+b+1=0\\ \Rightarrow a+b=-1.\end{array}$