View source for 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:

<math>a x^2 + b x + c</math>

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

<math>\left(\alpha x + \beta \right)^2</math>

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:

<math>\left(\alpha x + \beta \right)^2</math>

<math> = \alpha^2 x^2 + 2 \alpha \beta x + \beta^2</math>

<math> \equiv a x^2 + b x + c </math>

Identifying the coefficients we get:

<math> \alpha \equiv \sqrt{a} </math>

<math> \beta \equiv \sqrt{c} </math>

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

<math> b = 2 \alpha \beta = 2 \sqrt{a c} </math>

<math> b^2 = 4 a c</math>

==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 <math>\left(\alpha x + \beta \right)^2 = \alpha^2 \left(x + \frac{\beta}{\alpha} \right) \left(x + \frac{\beta}{\alpha} \right) </math>. (A double root is present at <math>x = -\frac{\beta}{\alpha}</math>)
[[Category:Mathematics]]