Solve equation $xy^2-(3x^2-4x+1)y+x^3-2x^2+x=0$ , in $\mathbb R$
This question was asked by another user but unfortunately was deleted when I was typing the answer. I did not want my time wasted, so I asked it again with new condition.
Solution:
Let $y=tx$ we get:
$(t^2-3t+1)x^3+2(2t-1)x^2+(1-t)x=0$
$x=0$
$(t^2-3t+1)x^2+2(2t-1)x+(1-t)=0$
$\Delta'=(2t-1)^2+(t-1)(t^2-3t+1)=k^2$
$\Delta'=t^3=k^2$
If $t=a^2$ then $t^3=(a^3)^2=k^2$
For $a∈\mathbb N$ we can have infinite solutions;
$(t, k)=(1, 1), (4, 8), (9, 27), \cdot\cdot\cdot$
$(t, k)=1, 1$ ⇒$(x, y)=(0, 0),(-2, -2) $
$(t, k)=(4, 8)$ ⇒ $(x, y)=(3, 12), (-\frac 15, -\frac 45)$
$(t, k) =(9, 27)$ ⇒ $(x, y)=(\frac 45, \frac{36}5),(\frac{-2} {11}, -\frac{18}{11})$
Any alternative solution?
Here is a plot by wolfram
Not sure why you put $y=ax$, since this is clearly a quadratic equation in $y$: $$xy^2-(3x^2-4x+1)y+x^3-2x^2+x=0$$
Its discriminant is $$D=(3x-1)^2(x-1)^2-4x^2(x-1)^2=(x-1)^3(5x-1)$$
So $y$ is defined for $x\in \left(-\infty,{1\over 5}\right]\cup [1,\infty)$ and $x\ne 0$: $$y={(x-1)\Big(3x-1\pm \sqrt{(x-1)(5x-1)}\Big)\over 2x}$$