Solving a quadratic equation in two variables: $x^2-4xy+3y^2=0$

Question

How can we solve this equation?

$$x^2-4xy+3y^2=0$$

How many values of $x$ and $y$ are possible? (There are two variables so it may not be possible to solve it, but how can we guess the number of real solutions, i.e., real values of $x$ and $y$?)

Answer 1

If $x^2 - 4yx + 3y^2 = 0$, we can view this as a quadratic equation in standard form with $a=1$, $b=-4y$, and $c=3y^2$; so, we can use the quadratic formula just as in the single variable case:

$$x = \frac{4y \pm \sqrt{16y^2 - 12y^2}}{2} = \frac{4y \pm \sqrt{4y^2}}{2} = \frac{4y \pm 2y}{2}$$

Tackling the $\pm$ as plus or minus, respectively, we have: $x = 3y$ or $x = y$.

Answer 2

Note $$x^2-4xy+3y^2=(x-y)(x-3y).$$

We got infinitely many solutions: $$\{(x,x),(3x,x)|x\in\mathbb R\}$$

Answer 3

There are two variables so it is not possible to solve it

The equation is homogeneous, and dividing by $y^2$ (*) reduces it to a quadratic $\,t^2-4t+3=0\,$ where $\,t=x/y\,$. Solving the quadratic, either by factoring or by formula, gives $\,t \in \{1,3\}\,$, so the two families of solutions (given by $\,x = ty\,$) are $\,x=y\,$ and $\,x=3y\,$ for an arbitrary $\,y \in \mathbb{R}\,$.

(*) Dividing by $\,y\,$ assumes $\,y \ne 0\,$, which corresponds to $\,x=0\,$, so this particular solution is included in both sets of solutions found above.