If `(2b-7)(2b+7)=1`, what is the value of `2b^(2)`?

Options

A. `15`
B. `25`
C. `32`
D. `50`

Answer

To solve the equation \((2b-7)(2b+7) = 1\) and find the value of \(2b^2\), we can follow these steps:

### Step 1: Recognize the difference of squares
The expression \((2b-7)(2b+7)\) can be recognized as a difference of squares. We can rewrite it using the identity \(a^2 - b^2 = (a-b)(a+b)\).

### Step 2: Apply the difference of squares formula
Let \(a = 2b\) and \(b = 7\). Thus, we have:
\[ (2b-7)(2b+7) = (2b)^2 - 7^2 \]
So, we can rewrite the equation as:
\[ (2b)^2 - 7^2 = 1 \]

### Step 3: Simplify the equation
Now, substituting \(7^2 = 49\):
\[ (2b)^2 - 49 = 1 \]

### Step 4: Rearrange the equation
To isolate \((2b)^2\), we add \(49\) to both sides:
\[ (2b)^2 = 1 + 49 \]
\[ (2b)^2 = 50 \]

### Step 5: Find \(2b^2\)
Now, we need to find \(2b^2\). We know that:
\[ (2b)^2 = 4b^2 \]
Thus, we can write:
\[ 4b^2 = 50 \]

### Step 6: Divide to find \(2b^2\)
To find \(2b^2\), we divide both sides of the equation by \(2\):
\[ 2b^2 = \frac{50}{2} \]
\[ 2b^2 = 25 \]

### Final Answer
The value of \(2b^2\) is \(25\).

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