Question

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

Answer 1

Let $\alpha$ be the common root for both equations then
$(\alpha {)}^2+a\alpha +b=0\dots (1)$
$(\alpha {)}^2+b\alpha +a=0\dots (2)$
Subtracting $\left(2\right)$ from $\left(1\right)$ we get,
$(a-b)\alpha +(b-a)=0$
Hence, $\alpha =1$
Substituting value of $\alpha$ in $\left(1\right)$ we get
$1+a+b=0$
$a+b=-1$.