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 the given formula

We have the formula for the number of diagonals \( d=\frac{n(n - 3)}{2} \). First, we want to isolate \( n \). Multiply both sides of the equation by 2 to get rid of the denominator:
\( 2d=n(n - 3) \)

Step2: Expand the right - hand side

Expand \( n(n - 3) \) to get a quadratic equation in terms of \( n \):
\( 2d=n^{2}-3n \)
Rearrange the equation to the standard quadratic form \( ax^{2}+bx + c = 0 \) (here \( x=n \)):
\( n^{2}-3n-2d = 0 \)

Step3: Use the quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the solutions for \( x \) are given by \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \). In our equation \( n^{2}-3n - 2d=0 \), we have \( a = 1 \), \( b=-3 \), and \( c=-2d \).
Substitute these values into the quadratic formula:
\( n=\frac{-(-3)\pm\sqrt{(-3)^{2}-4\times1\times(-2d)}}{2\times1}=\frac{3\pm\sqrt{9 + 8d}}{2} \)
We can rewrite \( \sqrt{9 + 8d} \) as \( 2\sqrt{2d+\frac{9}{4}} \) (because \( 9 + 8d=4(2d+\frac{9}{4}) \), so \( \sqrt{9 + 8d}=\sqrt{4(2d+\frac{9}{4})}=2\sqrt{2d+\frac{9}{4}} \))
Then \( n=\frac{3\pm2\sqrt{2d+\frac{9}{4}}}{2}=\frac{3}{2}\pm\sqrt{2d+\frac{9}{4}} \)
Since the number of sides \( n \) must be positive, we take the positive root:
\( n=\sqrt{2d+\frac{9}{4}}+\frac{3}{2} \)

Answer:

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