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...?

  • 2
    $\begingroup$ In this specific case, you can factor $(x-y)$ if you compute the difference between the two equations. $\endgroup$ – 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




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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.