The greatest positive integral value of x for which 200 - x(10 + x) is positive is

Options

A. 9
B. 10
C. 8
D. 11

Answer

To solve the problem, we need to find the greatest positive integral value of \( x \) for which the expression \( 200 - x(10 + x) \) is positive.

### Step-by-step Solution:

1. **Set Up the Inequality**:
We start with the expression:
\[ 200 - x(10 + x) > 0 \]
This can be rewritten as:
\[ 200 - 10x - x^2 > 0 \]

2. **Rearranging the Inequality**:
Rearranging gives us:
\[ -x^2 - 10x + 200 > 0 \]
To make it easier to work with, we can multiply the entire inequality by -1 (remember to flip the inequality sign):
\[ x^2 + 10x - 200 < 0 \]

3. **Finding the Roots of the Quadratic**:
We need to find the roots of the quadratic equation:
\[ x^2 + 10x - 200 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 10 \), and \( c = -200 \):
\[ x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 1 \cdot (-200)}}{2 \cdot 1} \]
\[ x = \frac{-10 \pm \sqrt{100 + 800}}{2} \]
\[ x = \frac{-10 \pm \sqrt{900}}{2} \]
\[ x = \frac{-10 \pm 30}{2} \]

4. **Calculating the Roots**:
This gives us two possible solutions:
\[ x = \frac{20}{2} = 10 \quad \text{and} \quad x = \frac{-40}{2} = -20 \]

5. **Analyzing the Quadratic**:
The roots of the quadratic \( x^2 + 10x - 200 \) are \( x = 10 \) and \( x = -20 \). The quadratic opens upwards (since the coefficient of \( x^2 \) is positive), which means it will be negative between the roots:
\[ -20 < x < 10 \]

6. **Finding the Greatest Positive Integral Value**:
Since we are looking for the greatest positive integral value of \( x \) that satisfies the inequality \( -20 < x < 10 \), the largest integer in this range is:
\[ x = 9 \]

### Conclusion:
The greatest positive integral value of \( x \) for which \( 200 - x(10 + x) \) is positive is:
\[ \boxed{9} \]