Given that $\alpha = \sin 36^\circ$, we need to determine which equation it is a root of. We start with the known relationship for $\cos 72^\circ$: $\cos 72^\circ = \frac{\sqrt{5}-1}{4}$ Using the double-angle formula for cosine: $\cos 72^\circ = 1 - 2 \sin^2 36^\circ$ Substitute $\alpha$ for $\sin 36^\circ$: $1 - 2\alpha^2 = \frac{\sqrt{5}-1}{4}$ Multiply both sides by 4: $4 - 8\alpha^2 = \sqrt{5} - 1$ Add 1 to both sides: $5 - 8\alpha^2 = \sqrt{5}$ Square both sides to eliminate the radical: $(5 - 8\alpha^2)^2 = 5$ Expand the left side: $25 + 64\alpha^4 - 80\alpha^2 = 5$ Simplify by subtracting 5 from both sides: $64\alpha^4 - 80\alpha^2 + 20 = 0$ Divide the entire equation by 4: $16\alpha^4 - 20\alpha^2 + 5 = 0$ Thus, the equation $16\alpha^4 - 20\alpha^2 + 5 = 0$ is the one for which $\alpha = \sin 36^\circ$ is a root.