If the arithmetic mean of two numbers is 5 and geometric mean is 4 then the numbers are

Options

A. 4,6
B. 4,7
C. 3,8
D. 2,8

Answer

To solve the problem, we need to find two numbers \( a \) and \( b \) given that their arithmetic mean is 5 and their geometric mean is 4.

### Step-by-Step Solution:

1. **Understanding the Means:**
- The arithmetic mean (AM) of two numbers \( a \) and \( b \) is given by:
\[ \text{AM} = \frac{a + b}{2} \]
- The geometric mean (GM) of two numbers \( a \) and \( b \) is given by:
\[ \text{GM} = \sqrt{ab} \]

2. **Setting Up the Equations:**
- From the problem, we know:
\[ \frac{a + b}{2} = 5 \]
Multiplying both sides by 2, we get:
\[ a + b = 10 \quad \text{(Equation 1)} \]
- We also know:
\[ \sqrt{ab} = 4 \]
Squaring both sides, we get:
\[ ab = 16 \quad \text{(Equation 2)} \]

3. **Using the Equations:**
- Now we have two equations:
1. \( a + b = 10 \)
2. \( ab = 16 \)

4. **Finding the Numbers:**
- We can use these equations to form a quadratic equation. Let \( a \) and \( b \) be the roots of the equation:
\[ x^2 - (a+b)x + ab = 0 \]
- Substituting the values from our equations:
\[ x^2 - 10x + 16 = 0 \]

5. **Solving the Quadratic Equation:**
- We can solve this using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 1 \), \( b = -10 \), and \( c = 16 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \]
\[ x = \frac{10 \pm \sqrt{100 - 64}}{2} \]
\[ x = \frac{10 \pm \sqrt{36}}{2} \]
\[ x = \frac{10 \pm 6}{2} \]

6. **Calculating the Roots:**
- This gives us two possible values:
\[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{4}{2} = 2 \]

7. **Final Numbers:**
- Therefore, the two numbers are \( a = 8 \) and \( b = 2 \).

### Conclusion:
The two numbers are \( 8 \) and \( 2 \).