TITLE: Quadratic solutions puzzle
QUESTION [1 upvotes]: 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?

REPLY [0 votes]: 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.
{"set_name": "stack_exchange", "score": 1, "question_id": 1279477}