Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$b=\frac{-\sqrt{a^2-4 b}-a}2$$
$$a=\frac{-\sqrt{a^2-4 b}-a}2$$
$$b=\frac{\sqrt{a^2-4 b}-a}2$$
The only solution for all of these is $a=b=0$, but is this right?

Let's take $a=\frac{\sqrt{a^2-4 b}-a}2$ for example:
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$2a=\sqrt{a^2-4 b}-a$$
$$3a=\sqrt{a^2-4 b}$$
$$9a^2=a^2-4 b$$
$$8a^2=-4 b$$
$$b=-2 a^2$$
Now you can take that $b$ and substitute in $b=\frac{\sqrt{a^2-4 b}-a}2$ to find a.
