If `(x + 7954 xx 7956)` be a square number, then the value of x is.
यदि `(x + 7954 xx 7956)` एक वर्ग संख्या हो तो 'x' का मान क्या होगा?

Options

A. A)1
B. B)16
C. C)9
D. D)4

Explanation

To solve the problem, we need to determine the value of \( x \) such that \( (x + 7954 \times 7956) \) is a perfect square.

### Step-by-Step Solution:

1. **Calculate \( 7954 \times 7956 \)**:
We can use the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \). Here, we can set \( a = 7955 \) (the average of 7954 and 7956) and \( b = 1 \) (the difference from the average).
\[ 7954 \times 7956 = (7955 - 1)(7955 + 1) = 7955^2 - 1^2 = 7955^2 - 1 \]

2. **Calculate \( 7955^2 \)**:
We can compute \( 7955^2 \) directly:
\[ 7955^2 = 63280225 \]
Therefore,
\[ 7954 \times 7956 = 63280225 - 1 = 63280224 \]

3. **Set up the equation**:
We want \( x + 63280224 \) to be a perfect square:
\[ x + 63280224 = n^2 \quad \text{for some integer } n \]
Rearranging gives:
\[ x = n^2 - 63280224 \]

4. **Find \( n^2 \) that is just greater than \( 63280224 \)**:
We need to find the smallest integer \( n \) such that \( n^2 > 63280224 \). Taking the square root:
\[ n > \sqrt{63280224} \approx 7960 \]
The smallest integer \( n \) is 7961.

5. **Calculate \( x \)**:
Now we compute \( x \):
\[ x = 7961^2 - 63280224 \]
First, calculate \( 7961^2 \):
\[ 7961^2 = 63299221 \]
Now substituting back:
\[ x = 63299221 - 63280224 = 18997 \]

6. **Check for perfect squares**:
Next, we check if \( x + 63280224 \) is a perfect square:
\[ 18997 + 63280224 = 63299221 \]
Since \( 63299221 = 7961^2 \), it is indeed a perfect square.

7. **Final answer**:
The value of \( x \) that makes \( (x + 7954 \times 7956) \) a perfect square is \( 18997 \).