Find the positive integer whose square exceeds ten times itself by 96.

Options

A. 6
B. 16
C. 9
D. none of these

Answer

To find the positive integer whose square exceeds ten times itself by 96, we can follow these steps:

### Step 1: Define the variable
Let the positive integer be represented by \( x \).

### Step 2: Set up the equation
According to the problem, the square of the integer exceeds ten times the integer by 96. This can be expressed mathematically as:
\[ x^2 = 10x + 96 \]

### Step 3: Rearrange the equation
To form a standard quadratic equation, we rearrange the equation:
\[ x^2 - 10x - 96 = 0 \]

### Step 4: Factor the quadratic equation
Now we need to factor the quadratic equation \( x^2 - 10x - 96 = 0 \). We look for two numbers that multiply to \(-96\) and add up to \(-10\). The numbers \(-16\) and \(6\) fit this requirement:
\[ x^2 - 16x + 6x - 96 = 0 \]
Grouping the terms gives:
\[ (x - 16)(x + 6) = 0 \]

### Step 5: Solve for \( x \)
Setting each factor to zero gives us the possible solutions:
\[ x - 16 = 0 \quad \Rightarrow \quad x = 16 \]
\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \]

### Step 6: Identify the positive integer
Since we are looking for a positive integer, we discard \( x = -6 \) and keep:
\[ x = 16 \]

### Conclusion
The positive integer whose square exceeds ten times itself by 96 is:
\[ \boxed{16} \]

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