Find all $a, b \in \Bbb R$, ($b\ne0)$, such that the roots of
$$x^2+ax+a=b$$ $$x^2+ax+a=-b$$
are 4 consecutive numbers.
We have: $$x^2+ax+a-b=0$$ $$x^2+ax+a+b=0$$
$x_1, x_2$ – roots of first equation; $x_3, x_4$ – roots of second equation.
By logic I wrote: $x_2<x_4<x_3<x_1$ (Is there a way to prove this?)
Using Vieta's formulas: $$x_1+x_2+x_3+x_4=2a$$ and because they are consecutive numbers we have $$x=\frac{a-3}2$$
$$\frac{a-3}2=x_2<\frac{a-1}2=x_4<\frac{a+1}2=x_3<\frac{a+3}2=x_1$$
Using Vieta's formulas again: $$\frac{a^2-9}4=a-b,\quad \frac{a^2-1}4=a+b$$
$$\implies b=-1$$
$$\implies a_{1/2}=2\pm\frac{\sqrt{58}}2,\quad a_3=5,\quad a_4=-1$$
My question is are these all the solutions? Or are there more?
Best Answer
You made a mistake while solving $b$. Specifically, $b=1$. As such, $$a^2-4a-5 = (a-5)(a+1)=0 \implies a = -1,5.$$
Regarding $x_2 < x_4 < x_3 < x_1$, you could argue by using the following argument.
Note that one of the roots is the minimum among them. Say that is $x_2$. Since $$x_1 + x_2 = a = x_3 + x_4,$$ we have $$x_1 = x_3 + (x_4 - x_2) \implies x_1 > x_3$$ because $x_4 > x_2$. Likewise, $x_1 > x_4$.
You are looking for integer solutions. In that case, $a$ and $b$ must be integers