The HM of two numbers is 4. If their arithmetic mean A and geometric mean G satisfy the relation `2A + G^(2) = 27`, then the numbers are

Options

A. 2, 6
B. 3, 6
C. 1, 3
D. 1, 2

Answer

To solve the problem step by step, we will use the definitions of harmonic mean (HM), arithmetic mean (A), and geometric mean (G) and the given conditions.

### Step 1: Write down the definitions and given information.
Let the two numbers be \( a \) and \( b \).

- The harmonic mean (HM) of two numbers is given by:
\[ HM = \frac{2ab}{a + b} \]
- We are given that \( HM = 4 \). Therefore, we can write:
\[ \frac{2ab}{a + b} = 4 \tag{1} \]

### Step 2: Express the arithmetic mean (A) and geometric mean (G).
- The arithmetic mean (A) is given by:
\[ A = \frac{a + b}{2} \]
- The geometric mean (G) is given by:
\[ G = \sqrt{ab} \]

### Step 3: Use the relationship between A, G, and the given equation.
We are given the equation:
\[ 2A + G^2 = 27 \tag{2} \]
Substituting the expressions for A and G into equation (2):
\[ 2\left(\frac{a + b}{2}\right) + (\sqrt{ab})^2 = 27 \]
This simplifies to:
\[ a + b + ab = 27 \tag{3} \]

### Step 4: Solve the equations (1) and (3).
From equation (1):
\[ 2ab = 4(a + b) \implies ab = 2(a + b) \tag{4} \]
Now we have two equations (3) and (4):
1. \( a + b + ab = 27 \) (from equation 3)
2. \( ab = 2(a + b) \) (from equation 4)

### Step 5: Substitute equation (4) into equation (3).
Substituting \( ab \) from equation (4) into equation (3):
\[ a + b + 2(a + b) = 27 \]
This simplifies to:
\[ 3(a + b) = 27 \]
Thus,
\[ a + b = 9 \tag{5} \]

### Step 6: Substitute equation (5) back into equation (4).
Now substituting \( a + b = 9 \) into equation (4):
\[ ab = 2(9) = 18 \tag{6} \]

### Step 7: Solve for a and b using equations (5) and (6).
Now we have a system of equations:
1. \( a + b = 9 \)
2. \( ab = 18 \)

These can be treated as the roots of the quadratic equation:
\[ x^2 - (a + b)x + ab = 0 \]
Substituting the values from equations (5) and (6):
\[ x^2 - 9x + 18 = 0 \]

### Step 8: Solve the quadratic equation.
To find the roots, we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -9, c = 18 \):
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} \]
\[ x = \frac{9 \pm \sqrt{81 - 72}}{2} \]
\[ x = \frac{9 \pm \sqrt{9}}{2} \]
\[ x = \frac{9 \pm 3}{2} \]
This gives us:
\[ x = \frac{12}{2} = 6 \quad \text{or} \quad x = \frac{6}{2} = 3 \]

### Conclusion: The two numbers are
Thus, the two numbers are \( 6 \) and \( 3 \).

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