Identify the conic section $$$- x^{2} + x = \frac{1}{16}$$$

Identify and find the properties of the conic section $$$- x^{2} + x = \frac{1}{16}$$$.

The general equation of a conic section is $$$A x^{2} + B x y + C y^{2} + D x + E y + F = 0$$$.

In our case, $$$A = 1$$$, $$$B = 0$$$, $$$C = 0$$$, $$$D = -1$$$, $$$E = 0$$$, $$$F = \frac{1}{16}$$$.

The discriminant of the conic section is $$$\Delta = 4 A C F - A E^{2} - B^{2} F + B D E - C D^{2} = 0$$$.

Next, $$$B^{2} - 4 A C = 0$$$.

Since $$$\Delta = 0$$$, this is the degenerated conic section.

Since $$$B^{2} - 4 A C = 0$$$, the equation represents two parallel lines.

$$$- x^{2} + x = \frac{1}{16}$$$A represents a pair of the lines $$$x = - \frac{-2 + \sqrt{3}}{4}$$$, $$$x = \frac{\sqrt{3} + 2}{4}$$$A.

General form: $$$x^{2} - x + \frac{1}{16} = 0$$$A.

Factored form: $$$\left(4 x - 2 - \sqrt{3}\right) \left(4 x - 2 + \sqrt{3}\right) = 0$$$A.

Graph: see the graphing calculator.