Mathematics - Quadratic Equation Question with Solution | TestHub

Let $\alpha$ be the common root. Then it must satisfy both equations: ${\alpha }^2+a\alpha +b=0$ (1) and ${\alpha }^2+b\alpha +a=0$ (2). Subtracting equation (2) from equation (1), we get $(a-b)\alpha +(b-a)=0$, which simplifies to $(a-b)\alpha -(a-b)=0$. Factoring out $(a-b)$, we have $(a-b)(\alpha -1)=0$. Since $a\ne b$, it implies $a-b\ne 0$. Therefore, we must have $\alpha -1=0$, which means $\alpha =1$. Substituting $\alpha =1$ into equation (1), we get $1^2+a(1)+b=0$, so $1+a+b=0$. Thus, $a+b=-1$.