Mathematics - Quadratic Equation Question with Solution | TestHub

Let $\alpha$ be the common root. Then ${\alpha }^2+a\alpha +b=0$ and ${\alpha }^2+b\alpha +a=0$. Subtracting the second equation from the first, we get $(a-b)\alpha +(b-a)=0$. This simplifies to $(a-b)\alpha -(a-b)=0$. Since $a\ne b$, we can divide by $(a-b)$, which gives $\alpha -1=0$, so $\alpha =1$. Substituting $\alpha =1$ into either equation, for example, the first one: $1^2+a(1)+b=0$, which means $1+a+b=0$. Therefore, $a+b=-1$.