Reference request: Transformations under which the discriminant is invariant

Question

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?

Answer 1

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.

Answer 2

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 $Q(x)$.