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 $y$?)
[Math] Solving a quadratic equation in two variables: $x^2-4xy+3y^2=0$
algebra-precalculusfactoringquadratics
Related Solutions
Well. . .
There's several problems here.
First, notice: $$(x+d)(x-e)=x^2-ex+dx-ed=x^2+(d-e)x-ed$$ Where in the world are you going to get a leading coefficient? In other words, this method doesn't suffice without some manipulations of the standard form. That manipulation is: $$\frac{1}{a}\left(ax^2+bx+c\right)=x^2+px+q \quad \quad p=\frac{b}{a},q=\frac{c}{a}$$ The second issue is that it can't be generally figured out what $d$ and $e$ are. $d$, by definition, is the opposite of the first root of the equation. $e$, by definition, is the second root. However, the relationships between $p$ and $q$ and $d$ and $e$ are: $$d-e=p \quad -ed=q$$ This information doesn't allow us to determine $d$ and $e$ in terms of the coefficients $p$ and $q$. There's also the common sensical fact, that like I said, you have to know the roots to know $e$ and $d$.
But the biggest issue is what you want. You cannot get rid of the imaginary solutions. They will always be solutions to particular quadratics. It is impossible to "get rid of them." For example, notice: $$x^2+1=0$$ There are no solutions to this equation in $\mathbb{R}$. The only solutions are in $\mathbb{C}$. They are $i$ and $-i$. Now, let me show why they have to be those particular quantities in a very simply way: $$\begin{align} x^2+1&=(x-r_1)(x-r_2)\\ &=(x-i)(x+i)\\ &=x(x+i)-i(x+i)\\ &=x^2+ix-ix-i^2\\ &=x^2-(-1)\\ &=x^2+1 \end{align}$$ Do you see how those solutions have to be $i$ and $-i$? This applies to all quadratics. Their roots are whatever makes the following relation true and the fundamental relation (that is, that the quadratic is equal to $0$) true. I hope this makes sense.
Addendum: It may help to see precisely where "that formula" comes from. It is derived directly from the standard form equation.
First, make this observation: $$(x-v)^2=x^2-2vx+v^2$$ You can turn the manipulated form into a square: $$\begin{align} x^2+px+q&=0\\ x^2+px&=-q\\ x^2+px+\left(\frac{p}{2}\right)^2&=-q+\left(\frac{p}{2}\right)^2\\ \left(x+\frac{p}{2}\right)^2&=-q+\left(\frac{p}{2}\right)^2=\frac{-4q+p^2}{4}\\ x+\frac{p}{2}&=\pm \sqrt{\frac{-4q+p^2}{4}}=\frac{1}{2}\sqrt{-4q+p^2}\\ x&=\pm \frac{1}{2}\sqrt{-4q+p^2}-\frac{p}{2}=\frac{-p\pm\sqrt{-4q+p^2}}{2} \end{align}$$ Substituting back in for $p$ and $q$ gives us: $$\begin{align} x&=\frac{-\frac{b}{a}\pm \sqrt{-4\frac{c}{a}+\left(\frac{b}{a}\right)^2}}{2}\\ &=\frac{-b \pm a\sqrt{-4\frac{c}{a}+\frac{b^2}{a^2}}}{2a}\\ &=\frac{-b \pm a\sqrt{\frac{b^2-4ac}{a^2}}}{2a}\\ &=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align}$$
Also, André Nicolas provides an even more simple derivation of the quadratic formula in this question: Why can ALL quadratic equations be solved by the quadratic formula?
Since you have two points for each line, we can find the formula for the unique line (assuming the points are distinct). For the first line, defined by $(x_1,y_1)$ and $(x_2,y_2)$, the slope is "rise-over-run", i.e. $\frac {y_2 - y_1} {x_2 - x_1}$, thus any point $x$ and $y$ on this line must have the same slope, so $$\frac {y - y_1} {x - x_1}= \frac {y_2 - y_1} {x_2 - x_1}\implies y=\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1\,.$$For the second line, we get the analogous result:
$$y=\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,.$$
Now we just have to solve these two equations, since we are looking for the point of intersection where the $x$ and $y$ variables for each line are equivalent, we set the $y's$ equal to each other and arrive at the equation:
$$\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1=\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,,$$ which we can simplify to $$\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]x= \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\,.$$
Now if $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]\ne 0$ we can divide to find a specific formula for $x$: $$x=\frac 1 {\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)}\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\right]\,.$$ Now, finding $y$ will be easy - we just "back substitute" $x$ into either equation for $y$, say the first one:
$$y=\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)\left (\frac 1 {\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)}\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1-y_1 - \left ( \frac {y_4 - y_3} {x_4 - x_3}\right)x_3+y_3\right] \right) - \left ( \frac {y_2 - y_1} {x_2 - x_1}\right)x_1+y_1\,.$$
This will always work, unless we have that $\left[\left ( \frac {y_2 - y_1} {x_2 - x_1}\right)-\left ( \frac {y_4 - y_3} {x_4 - x_3}\right)\right]=0$. In that case, you either have coincident lines or parallel lines - to determine which, just check the y-intercept of each line (we already know the slopes are equal because of the above condition). If the y-intercepts of each line are equal to each other, then the lines are coincident (so there are infinitely many "points of intersection"). If not, they are parallel (and there are no points of intersection).
Best Answer
Note $$x^2-4xy+3y^2=(x-y)(x-3y).$$
We got infinitely many solutions: $$\{(x,x),(3x,x)|x\in\mathbb R\}$$