Show that if A and G .are A.M. and G.M. between two positive numbers , then the numbers are `A+- sqrt(A^(2) -G^(2))`

Answer

To show that if \( A \) and \( G \) are the arithmetic mean (A.M.) and geometric mean (G.M.) between two positive numbers \( a \) and \( b \), then the numbers can be expressed as \( A \pm \sqrt{A^2 - G^2} \), we can follow these steps:

### Step 1: Define the means
Let \( A \) be the arithmetic mean and \( G \) be the geometric mean of the two positive numbers \( a \) and \( b \).

\[ A = \frac{a + b}{2} \]
\[ G = \sqrt{ab} \]

### Step 2: Express \( b \) in terms of \( a \)
From the definition of \( A \), we can express \( b \) in terms of \( a \):

\[ b = 2A - a \]

### Step 3: Substitute \( b \) into the expression for \( G \)
Now substitute \( b \) into the expression for \( G \):

\[ G = \sqrt{a(2A - a)} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives us:

\[ G^2 = a(2A - a) \]

### Step 5: Rearrange the equation
Rearranging the equation leads to:

\[ G^2 = 2aA - a^2 \]
\[ a^2 - 2aA + G^2 = 0 \]

### Step 6: Recognize the quadratic form
This is a quadratic equation in terms of \( a \):

\[ a^2 - 2Aa + G^2 = 0 \]

### Step 7: Apply the quadratic formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can find \( a \):

Here, \( b = -2A \), \( a = 1 \), and \( c = G^2 \):

\[ a = \frac{2A \pm \sqrt{(-2A)^2 - 4 \cdot 1 \cdot G^2}}{2 \cdot 1} \]
\[ a = \frac{2A \pm \sqrt{4A^2 - 4G^2}}{2} \]
\[ a = A \pm \sqrt{A^2 - G^2} \]

### Step 8: Find \( b \)
Now, substituting back to find \( b \):

\[ b = 2A - a = 2A - (A \pm \sqrt{A^2 - G^2}) = A \mp \sqrt{A^2 - G^2} \]

### Conclusion
Thus, we have shown that the two numbers can be expressed as:

\[ a = A \pm \sqrt{A^2 - G^2} \]
\[ b = A \mp \sqrt{A^2 - G^2} \]