Converting a 2nd degree polynomial to a perfect square

Ok, I honestly couldn't come up with a better name for this lesson.

Let's do it:

We have a second degree polynomial:

$a{x}^2+bx+c$

and we want to find a change of variable that transforms it to a perfect square:

${(\alpha x+\beta )}^2$

Of course, for the transformation to exist, the polynomial's coefficients must have 1 degree of freedom! We will assume it is so. Expanding the expression we get:

${(\alpha x+\beta )}^2$

$={\alpha }^2{x}^2+2\alpha \beta x+{\beta }^2$

$\equiv a{x}^2+bx+c$

Identifying the coefficients we get:

$\alpha \equiv \sqrt{a}$

$\beta \equiv \sqrt{c}$

And the condition (for which we required the degree of freedom):

$b=2\alpha \beta =2\sqrt{ac}$

$b^2=4ac$

Further interpretation of the condition obtained

Notice that the condition is equivalent to saying the determinant of the polynomial is equal to 0. When a polynomial's determinant was 0, the polynomial has a double root. Note this is just what we asked for when equating it to ${(\alpha x+\beta )}^2={\alpha }^2(x+\frac{\beta }{\alpha })(x+\frac{\beta }{\alpha })$. (A double root is present at $x=-\frac{\beta }{\alpha }$)