The arithmetic mean between two numbers is A and the geometric mean is G. Then these numbers are:

Options

A. `sqrt(A^2 -G^2 pmA)`
B. `1/2(sqrt(A^2 -G^2) +A)`
C. `A pm sqrt(A^2 -G^2)`
D. `2(A pm sqrt(A^2-G^2))`

Answer

To find the two numbers given their arithmetic mean \( A \) and geometric mean \( G \), we can follow these steps:

### Step 1: Define the Variables
Let the two numbers be \( x \) and \( y \).

### Step 2: Set Up the Equations
The arithmetic mean of \( x \) and \( y \) is given by:
\[ \frac{x + y}{2} = A \]
Multiplying both sides by 2 gives:
\[ x + y = 2A \quad \text{(Equation 1)} \]

The geometric mean of \( x \) and \( y \) is given by:
\[ \sqrt{xy} = G \]
Squaring both sides gives:
\[ xy = G^2 \quad \text{(Equation 2)} \]

### Step 3: Express One Variable in Terms of the Other
From Equation 2, we can express \( y \) in terms of \( x \):
\[ y = \frac{G^2}{x} \]

### Step 4: Substitute into the Arithmetic Mean Equation
Substituting \( y \) from the above equation into Equation 1:
\[ x + \frac{G^2}{x} = 2A \]
Multiplying through by \( x \) to eliminate the fraction gives:
\[ x^2 + G^2 = 2Ax \]

### Step 5: Rearrange into a Quadratic Equation
Rearranging the equation gives:
\[ x^2 - 2Ax + G^2 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = -2A \), and \( c = G^2 \).
\[ x = \frac{2A \pm \sqrt{(-2A)^2 - 4 \cdot 1 \cdot G^2}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} \]
\[ x = A \pm \sqrt{A^2 - G^2} \]

### Step 7: Find the Values of \( y \)
Using \( y = \frac{G^2}{x} \), we can find the corresponding values of \( y \):
1. For \( x = A + \sqrt{A^2 - G^2} \):
\[ y = \frac{G^2}{A + \sqrt{A^2 - G^2}} \]
2. For \( x = A - \sqrt{A^2 - G^2} \):
\[ y = \frac{G^2}{A - \sqrt{A^2 - G^2}} \]

### Final Result
Thus, the two numbers are:
\[ x = A + \sqrt{A^2 - G^2} \quad \text{and} \quad y = A - \sqrt{A^2 - G^2} \]