# 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:

${\displaystyle ax^{2}+bx+c}$

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

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

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:

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

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

${\displaystyle \equiv ax^{2}+bx+c}$

Identifying the coefficients we get:

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

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

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

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

${\displaystyle b^{2}=4ac}$

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