Subtracting both equations $$x(a-b)+b-a=0$$ $$(x-1)(a-b)=0$$ Since $a\not = b$ $$x=1$$ Substitution of x gives $$1+a+b=0$$ which is contradictory to the question. What did I do wrong?
There is a solution for this question that proves the required condition satisfactorily, but I want to the know reason behind this contradiction.
You assume that both equations have the same solution. This is not stated in the problem. Only the difference of the two solutions to each equation is identical.
Hint:
You have wrongly assumed a common root, then only the subtraction makes sense.
If $p,q;(p\ge q)$ are the roots of $$ct^2+dt+e=0$$
$p-q=\dfrac{-d+\sqrt{d^2-4ce}-(-d-\sqrt{d^2-4ce)}}2=?$
We need $$\sqrt{a^2-4\cdot1\cdot b}=\sqrt{b^2-4\cdot1\cdot a}$$
Let $c$ be the difference between roots of $x^2+ax+b$ and $x^2+bx+a$
(in other words, let $x_{1,2}^2+ax_{1,2}+b=0$,
$x_{3,4}^2+bx_{3,4}+a=0$,
$x_1\le x_2$,
$x_3\le x_4$ and
$x_1-x_3=x_2-x_4=c$), so
$x^2+ax+b\equiv(x-c)^2+b(x-c)+a$
$\begin{cases}a=-2c+b\\b=c^2-bc+a\end{cases}$
$b=c^2-bc-2c+b$
$(c-b-2)c=0$,
$b=c-2$ or $c=0$,
$c\ne0$ as $x^2+ax+b$ and $x^2+bx+a$ are not the same, so $b=c-2$
$a=-2c+b=-c-2$
so $a+b=-4$, QED.
Let $x_1,x_2$ be the roots of the first equation and $y_1,y_2$ of the second.
By the Vieta's: $$\begin{cases}x_1+x_2=-a \\ x_1x_2=b\end{cases} \Rightarrow x_1^2+x_2^2=a^2-2b\\ \begin{cases}y_1+y_2=-b \\ y_1y_2=a\end{cases} \Rightarrow y_1^2+y_2^2=b^2-2a$$ The difference of roots is equal: $$x_1-x_2=y_1-y_2 \Rightarrow \\ x_1^2+x_2^2-2x_1x_2=y_1^2+y_2^2-2y_1y_2 \Rightarrow \\ a^2-4b=b^2-4a \Rightarrow \\ (a-b)(a+b)=-4(a-b) \Rightarrow \\ a+b=-4.$$ Note: $a-b\ne 0$.
