The difference between the corresponding roots of $x^2+ax+b=0$ and $x^2+bx+ a=0$ is same and $a\not = b$ then prove that $a+b+4=0$

Question

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.

Answer 1

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.

Answer 2

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 $$\sqrt{a^2-4\cdot1\cdot b}=\sqrt{b^2-4\cdot1\cdot a}$$

Answer 3

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.

Answer 4

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: $a-b\ne 0$.