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...?

• In this specific case, you can factor $(x-y)$ if you compute the difference between the two equations. – Gribouillis Aug 5 '17 at 6:57

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 $x/y=1$.

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:

1. $ax^2+bx+c = (\sqrt{a}x+\sqrt{c})^2$,
2. $ax^2+bx+c = (\sqrt{a}x-\sqrt{c})^2$,
3. $ax^2+bx+c = -(\sqrt{-a}x+\sqrt{-c})^2$, or
4. $ax^2+bx+c = -(\sqrt{-a}x-\sqrt{-c})^2$.

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.

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 $\frac{x}{y}=1$