Question

the number of diagonals in a polygon, d, in terms of the number of sides of a polygon, n, is given by the equation $d = \frac{n(n - 3)}{2}$. what equation shows the number of sides of a polygon in terms of the number of diagonals?
a. $n=sqrt{\frac{2d}{3}}$
b. $n=sqrt{2d + 3}$
c. $n=sqrt{2d}+3$
d. $n=sqrt{2d+\frac{9}{4}}+\frac{3}{2}$

Explanation:

Step1: Start with given formula

$d=\frac{n(n - 3)}{2}$

Step2: Multiply both sides by 2

$2d=n(n - 3)=n^{2}-3n$

Step3: Rearrange to quadratic - form

$n^{2}-3n - 2d=0$

Step4: Use quadratic formula $n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $an^{2}+bn + c = 0$

Here $a = 1$, $b=-3$, $c=-2d$. So $n=\frac{3\pm\sqrt{(-3)^{2}-4\times1\times(-2d)}}{2\times1}=\frac{3\pm\sqrt{9 + 8d}}{2}$
Since $n>0$, we consider the positive root. First, rewrite $\sqrt{9 + 8d}=\sqrt{4(2d+\frac{9}{4})}=2\sqrt{2d+\frac{9}{4}}$. Then $n=\frac{3 + 2\sqrt{2d+\frac{9}{4}}}{2}=\sqrt{2d+\frac{9}{4}}+\frac{3}{2}$

Answer:

D. $n=\sqrt{2d+\frac{9}{4}}+\frac{3}{2}$