If `p=sqrt(72-sqrt(72-sqrt(72-sqrt72-......oo)))` then find the value of `2p^(2)+1`

Options

A. `-129`
B. `-163`
C. 129
D. 163

Answer

To solve the equation \( p = \sqrt{72 - \sqrt{72 - \sqrt{72 - \sqrt{72 - \ldots}}}} \), we will first express the infinite nested radical in a simpler form.

### Step 1: Set up the equation
We start by recognizing that the expression inside the square root is the same as \( p \):
\[ p = \sqrt{72 - p} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ p^2 = 72 - p \]

### Step 3: Rearrange the equation
Next, we rearrange the equation to bring all terms to one side:
\[ p^2 + p - 72 = 0 \]

### Step 4: Solve the quadratic equation
Now we can use the quadratic formula to solve for \( p \):
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -72 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot (-72) = 1 + 288 = 289 \]

Now substituting back into the quadratic formula:
\[ p = \frac{-1 \pm \sqrt{289}}{2 \cdot 1} = \frac{-1 \pm 17}{2} \]

Calculating the two possible values for \( p \):
1. \( p = \frac{16}{2} = 8 \)
2. \( p = \frac{-18}{2} = -9 \) (not valid since \( p \) must be non-negative)

Thus, we have:
\[ p = 8 \]

### Step 5: Calculate \( 2p^2 + 1 \)
Now we substitute \( p \) back into the expression \( 2p^2 + 1 \):
\[ 2p^2 + 1 = 2(8^2) + 1 = 2(64) + 1 = 128 + 1 = 129 \]

### Final Answer
The value of \( 2p^2 + 1 \) is:
\[ \boxed{129} \]