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 equals StartFraction n left-parenthesis n minus 3 right-parenthesis over 2 EndFraction.



What equation shows the number of sides of a polygon in terms of the number of diagonals?

Answer options with 4 options
A.
n equals StartRoot StartFraction 2 d over 3 EndFraction EndRoot

B.
n equals StartRoot 2 d plus 3 EndRoot

C.
n equals StartRoot 2 d EndRoot plus 3

D.
n equals StartRoot 2 d plus StartFraction 9 over 4 EndFraction EndRoot plus StartFraction 3 over 2 EndFraction

Answers

GPT-4o mini
To find the number of sides of a polygon \( n \) in terms of the number of diagonals \( d \), we start from the given equation for the number of diagonals:

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

We can first rearrange this equation to express \( n \) in terms of \( d \).

1. Multiply both sides by 2 to eliminate the fraction:

\[
2d = n(n - 3)
\]

2. Expand the right-hand side:

\[
2d = n^2 - 3n
\]

3. Rearrange this into standard quadratic form:

\[
n^2 - 3n - 2d = 0
\]

Next, we will solve this quadratic equation using the quadratic formula:

\[
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation \( a = 1, b = -3, c = -2d \). Therefore:

\[
b^2 - 4ac = (-3)^2 - 4(1)(-2d) = 9 + 8d = 8d + 9
\]

Now substituting \( a, b, c \) into the quadratic formula gives:

\[
n = \frac{-(-3) \pm \sqrt{8d + 9}}{2(1)} = \frac{3 \pm \sqrt{8d + 9}}{2}
\]

Since \( n \) must be a positive quantity (the number of sides of a polygon), we only consider the positive root:

\[
n = \frac{3 + \sqrt{8d + 9}}{2}
\]

Now we need to simplify it further. We can express the right hand side with factoring out the coefficient for clarity:

Rewriting \( 8d + 9 \) as \( 2(4d + \frac{9}{2}) \):

\[
n = \frac{3}{2} + \frac{1}{2}\sqrt{8d + 9}
\]

However, it appears we need to express it similarly to the answer options provided.

Since we see that none of the answer options directly match our derived-form using \( \sqrt{8d + 9} \), we will verify the discrete options by straight substitution to find a corresponding polynomial.

From the given options, it can be concluded that:

The only expression that has a similar form to root-based and linear combinations is found by inspecting:

The closest valid structure for solving \( d \) against derived forms shall isolate \( n = \sqrt{2d + 3} \), signaling the probable answer based on what round main output values should register.

So, based indirectly on your options, the best answer option to select would mathematically filter down as:

**Option B: \( n = \sqrt{2d + 3} \)**

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