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?

You can do this by:
$$(x-a)(x-b)=x^2-(a+b)x+ab=x^2+ax+b$$
Equating constant terms gives $ab=b$ which means $a=1$ or $b=0$
Equating coefficients of $x$ gives $-a-b=a$ or equivalently $2a=-b$
