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:

$$a x^2 + b x + c$$

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

$$\left(\alpha x + \beta \right)^2$$

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:

$$\left(\alpha x + \beta \right)^2$$

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

$$ \equiv a x^2 + b x + c $$

Identifying the coefficients we get:

$$ \alpha \equiv \sqrt{a} $$

$$ \beta \equiv \sqrt{c} $$

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

$$ b = 2 \alpha \beta = 2 \sqrt{a c} $$

$$ b^2 = 4 a c$$

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 $$\left(\alpha x + \beta \right)^2 = \alpha^2 \left(x + \frac{\beta}{\alpha} \right) \left(x + \frac{\beta}{\alpha} \right) $$. (A double root is present at $$x = -\frac{\beta}{\alpha}$$)