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?
 A: Hint: Viete's Equations tell us that
$$\begin{cases}ab=b\\{}\\a+b=-a\end{cases}$$
There aren't that many possibilities...(and one of them is with $\;a,b\neq 0\;$)
A: 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 $2a=-b$
A: 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.
A: if $x=a$ is a solution to $x^2+ax+b=0$ then $b=-2a^2$
if $x=b$ is a solution  $b^2+ab+b=0$ so either $b=0$ or $a+b+1=0$
the second solution gives $=-2a^2+a+1=0$
this equation has solutions $a=1$ and $a=-\frac12$
so it looks like we can have $x^2+x-2$ with solutions of 1 and -2 
and $x^2 -\frac12 x -\frac 12$ which has $x=-\frac12$ as a solution.
so $(a,b) = (1,-2)$ and $(a,b) = (-\frac 12,-\frac 12)$ are both valid solutions.
