If the A.M. and G.M. between two numbers are respectively 17 and 8, find the numbers.

Answer

To find the two numbers given that their Arithmetic Mean (A.M.) is 17 and their Geometric Mean (G.M.) is 8, we can follow these steps:

### Step 1: Set up the equations for A.M. and G.M.
The formulas for the A.M. and G.M. of two numbers \( a \) and \( b \) are given by:
\[ \text{A.M.} = \frac{a + b}{2} \]
\[ \text{G.M.} = \sqrt{ab} \]
From the problem, we know:
\[ \frac{a + b}{2} = 17 \]
\[ \sqrt{ab} = 8 \]

### Step 2: Solve for \( a + b \) using A.M.
From the A.M. equation:
\[ a + b = 2 \times 17 = 34 \]

### Step 3: Solve for \( ab \) using G.M.
From the G.M. equation:
\[ ab = 8^2 = 64 \]

### Step 4: Substitute \( b \) in terms of \( a \)
Using the equation \( a + b = 34 \), we can express \( b \) in terms of \( a \):
\[ b = 34 - a \]

### Step 5: Substitute \( b \) in the product equation
Now substitute \( b \) into the product equation \( ab = 64 \):
\[ a(34 - a) = 64 \]
Expanding this gives:
\[ 34a - a^2 = 64 \]

### Step 6: Rearrange the equation
Rearranging the equation:
\[ -a^2 + 34a - 64 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ a^2 - 34a + 64 = 0 \]

### Step 7: Solve the quadratic equation
Now we can use the quadratic formula to solve for \( a \):
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -34, c = 64 \):
\[ a = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{34 \pm \sqrt{1156 - 256}}{2} \]
\[ = \frac{34 \pm \sqrt{900}}{2} \]
\[ = \frac{34 \pm 30}{2} \]
This gives us two possible values for \( a \):
1. \( a = \frac{64}{2} = 32 \)
2. \( a = \frac{4}{2} = 2 \)

### Step 8: Find corresponding values for \( b \)
Using \( b = 34 - a \):
1. If \( a = 32 \), then \( b = 34 - 32 = 2 \).
2. If \( a = 2 \), then \( b = 34 - 2 = 32 \).

### Conclusion
The two numbers are \( 2 \) and \( 32 \).

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