Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Appendix A - A.4 - Quadratic Equations - Exercises - Page 561: 35

Answer

$n = 6$

Work Step by Step

$D = \frac{n(n-3)}{2}$ $D = \frac{n^{2}-3n}{2}$ $9 = \frac{n^{2}-3n}{2}$ $18 = n^{2}-3n$ $n^{2} - 3n - 18 = 0$ Find two numbers that can be multiplied together to get $-18$ and can be added together to get $-3$. $= 3$ and $-6$ $n^{2} + 3n - 6n - 18 = 0$ $n(n + 3) - 6(n+3) = 0$ $(n-6)(n+3) = 0$ $n = 6, -3$ Since the number of sides cannot be a negative number, then $n = 6$.
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