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? \\(\bigcirc\\) a. $n = \sqrt{\frac{2d}{3}}$ \\(\bigcirc\\) b. $n = \sqrt{2d + 3}$ \\(\bigcirc\\) c. $n = \sqrt{2d} + 3$ \\(\bigcirc\\) d. $n = \sqrt{2d + \frac{9}{4}} + \frac{3}{2}$

Explanation:

Step1: Start with the given formula

We have \(d=\frac{n(n - 3)}{2}\). First, multiply both sides by 2 to get \(2d=n(n - 3)\). Expand the right - hand side: \(2d=n^{2}-3n\).

Step2: Complete the square

For the quadratic equation \(n^{2}-3n-2d = 0\), where \(a = 1\), \(b=-3\), \(c=-2d\). The formula for completing the square is \(n^{2}+bn=(n+\frac{b}{2})^{2}-\frac{b^{2}}{4}\).
For \(n^{2}-3n\), we have \(n^{2}-3n=(n-\frac{3}{2})^{2}-\frac{9}{4}\). So the equation \(n^{2}-3n-2d = 0\) can be rewritten as \((n-\frac{3}{2})^{2}-\frac{9}{4}-2d=0\).
Then \((n - \frac{3}{2})^{2}=2d+\frac{9}{4}\).

Step3: Solve for \(n\)

Take the square root of both sides: \(n-\frac{3}{2}=\sqrt{2d+\frac{9}{4}}\) (we consider the positive root since \(n>0\) for the number of sides of a polygon).
Add \(\frac{3}{2}\) to both sides: \(n=\sqrt{2d+\frac{9}{4}}+\frac{3}{2}\).

Answer:

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