# Converting a 2nd degree polynomial to a perfect square

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

Let's do it:

We have a second degree polynomial:

$ax^{2}+bx+c$ 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:

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