## Elementary Geometry for College Students (6th Edition)

We can find the number of vertices if a regular polygon has 8 diagonals: $\frac{n(n-3)}{2} = 8$ $n(n-3) = 16$ $n^2-3n- 16 = 0$ We can use the quadratic formula to find the solutions of the equation: $n = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $n = \frac{-(-3) \pm \sqrt{(-3)^2-(4)(1)(-16)}}{(2)(1)}$ $n = \frac{3 \pm \sqrt{9+64}}{2}$ $n = \frac{3 \pm \sqrt{73}}{2}$ $n = \frac{3 - \sqrt{73}}{2}~~$ or $~~n = \frac{3 + \sqrt{73}}{2}$ $n = -2.77~~$ or $~~n = 5.77$ Since $n$ is not a whole number, there is no regular polygon with 8 diagonals.