Reference request: Transformations under which the discriminant is invariant I had an occasion to think about this quadratic equation:
$$
ax^2 +bx(1-x) + c(1-x)^2 = 0.
$$
Its solution is
$$
x = \frac{2c-b\pm\sqrt{b^2-4ac\,}}{2(a-b+c)}.
$$
The thing under the radical is the same thing we were all taught in eighth or ninth grade. The mapping $(a,b,c)\mapsto (a-b+c, b-2c, c)$ doesn't change it. Is this an instance of the conclusion of some well known result?
 A: Let $P(x)=ax^2+bx(1-x)+c(1-x)^2$. Then 
\begin{align}
P(x)&=ax^2+bx(1-x)+c(1-x)^2\\
    &=x^2\big [a+b(\frac{1}{x}-1)+c(\frac{1}{x}-1)^2\big ]
\end{align}
Hence, $P(x)$ is obtained from $a+bx+cx^2$ by a projective transformation. Also, 
\begin{align}
a+bx+cx^2 &= x^2\big [a\frac{1}{x^2}+b\frac{1}{x}+c\big ]\\
\end{align}
and so $a+bx+cx^2$ can be obtained from $Q(x)=ax^2+bx+c$ by a projective transformation. It's well known that the discriminant is invariant under these projective transformations and so we have that the discriminant of $P(x)$ is equal to the discriminant of $Q(x)$. 
A: I don't see anything special here. Suppose that:
$$(a,b,c)\mapsto (a+x b+yc, b+zc, c)$$
Condition:
$$(b+zc)^2-4(a+xb+yc)c=b^2-4ac$$
leads to:
$$x=\frac z2,\quad y=\frac{z^2}{4}$$
You can easly construct infinitely many "nice" mappings, just pick even integer value for $z$:
$$(a,b,c)\mapsto (a+3b+9c, b+6c, c)$$
$$(a,b,c)\mapsto (a-4b+16c, b-8c, c)$$
Radical is also symmetrical with respect to $a,c$ so you can swap their roles and construct additional examples, like: 
$$(a,b,c)\mapsto (a, b-8a, c-4b+16a)$$
More general linear mappings are also possible but I don't find much value in investigating it further.
