Solving System of quadratic equations If $b²-4ac=0$ ($a \neq 0$,  and $a, b, c \in \mathbb {R}$) and  $x, y $ satisfy the system $$ax²+(b+3)x+c=3y$$ and $$ay²+(b+3)y+c=3x$$ then the value of $x/y$ is...?
 A: On adding the two equations we get
$$a(x^2+y^2)+b(x+y)+2c =0.$$
Which can be written as
$$(ax^2+bx+c)+(ay^2+by+c)=0.$$
Since $b^2-4ac=0$ (equal roots), therefore $ax^2+bx+c=a(x-\alpha)^2$, likewise $ay^2+by+c=a(y-\alpha)^2$.
Thus the above equation can be written as:
$$a[(x-\alpha)^2+(y-\alpha)^2]=0.$$
Since $a\neq 0$, therefore $x=y=\alpha$. Thus $x/y=1$.
A: Assuming $x,y \in \mathbb{R}$, then $x/y = 1$.
Your equations give
$$
  ax^2+bx+c = 3(y-x) = -3(x-y) = -(ay^2+by+c),
$$
so
$$ ax^2+bx+c = -(ay^2+by+c). $$
The assumption $b^2-4ac=0$, $a\neq 0$, means that one of the following must be true:


*

*$ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2$,

*$ax^2+bx+c = (\sqrt{a}x-\sqrt{c})^2$,

*$ax^2+bx+c = -(\sqrt{-a}x+\sqrt{-c})^2$, or

*$ax^2+bx+c = -(\sqrt{-a}x-\sqrt{-c})^2$.


For simplicity say we are in the first case. Then
$$
  (\sqrt{a}x+\sqrt{c})^2 = -(\sqrt{a}y+\sqrt{c})^2 .
$$
This can only happen if $\sqrt{a}x+\sqrt{c} = \sqrt{a}y+\sqrt{c} = 0$, and then $x=y=-\sqrt{c}/\sqrt{a}$. The other three cases are similar.
A: Let $u(v)=av^2+bv+c$. Since $b^2-4ac=0$, then $u=0$ has root $v=\frac{-b}{2a}$.
So $u(\frac{-b}{2a})=0$
If you reorder the equations above, you can write 
$ax^2+bx+c=3y-3x$
and
$ay^2+by+c=3x-3y$
Let $p(x)=u(x)-f(x,y)$ and $p(y)=u(y)+f(x,y)$ where $f(x,y)=3x-3y$
Adding the $p$'s together you get $p(x)+p(y)=u(x)+u(y)$
Entering in $\frac{-b}{2a}$ for both $x$ and $y$ gives
$2 p(\frac{-b}{2a})=0+0$
This means we have that $p(-\frac{b}{2a})=0$.
This means when you solve the above system of equations you get $x=y=\frac{-b}{2a}$
And $\frac{x}{y}=1$
