If `(4p+1)^(2)=81 and pgt0`, what is a possible value of p?

Answer

To solve the equation \((4p + 1)^2 = 81\) with the condition that \(p > 0\), we can follow these steps:

### Step 1: Take the square root of both sides
We start by taking the square root of both sides of the equation:

\[ 4p + 1 = \sqrt{81} \]

### Step 2: Simplify the square root
Since \(\sqrt{81} = 9\), we can write:

\[ 4p + 1 = 9 \quad \text{or} \quad 4p + 1 = -9 \]

### Step 3: Solve the first equation
Now, let's solve the first equation:

\[ 4p + 1 = 9 \]

Subtract 1 from both sides:

\[ 4p = 9 - 1 \]

\[ 4p = 8 \]

Now, divide by 4:

\[ p = \frac{8}{4} = 2 \]

### Step 4: Solve the second equation
Next, we solve the second equation:

\[ 4p + 1 = -9 \]

Subtract 1 from both sides:

\[ 4p = -9 - 1 \]

\[ 4p = -10 \]

Now, divide by 4:

\[ p = \frac{-10}{4} = -2.5 \]

### Step 5: Determine the valid solution
Since we are given that \(p > 0\), we discard \(p = -2.5\) as it does not satisfy the condition. Therefore, the only valid solution is:

\[ p = 2 \]

### Final Answer
The possible value of \(p\) is \(2\).

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