The AM of two observations is 25, and GM is 15. Find the two observations and their HM.

Given:
- Arithmetic Mean (AM) of two observations = 25
- Geometric Mean (GM) of two observations = 15

Let the two observations be \( x \) and \( y \).

### Step 1: Use the AM to form the first equation
The AM of \( x \) and \( y \) is given by:

\[
\frac{x + y}{2} = 25
\]

Multiplying both sides by 2:

\[
x + y = 50 \quad \text{(Equation 1)}
\]

### Step 2: Use the GM to form the second equation
The GM of \( x \) and \( y \) is given by:

\[
\sqrt{xy} = 15
\]

Squaring both sides:

\[
xy = 225 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations
We now have two equations:

\[
x + y = 50
\]
\[
xy = 225
\]

These equations can be solved using the quadratic formula. Consider the quadratic equation:

\[
t^2 - (x+y)t + xy = 0
\]

Substituting the values from Equations 1 and 2:

\[
t^2 - 50t + 225 = 0
\]

### Step 4: Solve the quadratic equation
To solve this, use the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation \( t^2 - 50t + 225 = 0 \):

\[
t = \frac{50 \pm \sqrt{50^2 - 4 \times 1 \times 225}}{2 \times 1}
\]
\[
t = \frac{50 \pm \sqrt{2500 - 900}}{2}
\]
\[
t = \frac{50 \pm \sqrt{1600}}{2}
\]
\[
t = \frac{50 \pm 40}{2}
\]

So, the two roots (observations) are:

\[
t_1 = \frac{50 + 40}{2} = 45
\]
\[
t_2 = \frac{50 - 40}{2} = 5
\]

Therefore, the two observations are \( x = 45 \) and \( y = 5 \).

### Step 5: Calculate the Harmonic Mean (HM)
The Harmonic Mean (HM) of \( x \) and \( y \) is given by:

\[
\text{HM} = \frac{2xy}{x + y}
\]

Substituting the values:

\[
\text{HM} = \frac{2 \times 45 \times 5}{45 + 5} = \frac{450}{50} = 9
\]

### Final Answer
- The two observations are **45** and **5**.
- The Harmonic Mean (HM) is **9**.

Would you like any more details on this solution, or do you have any further questions?

Here are 5 related questions you might find interesting:
1. How does the AM-GM-HM inequality relate to the values found here?
2. What are the conditions under which the AM equals the GM or HM?
3. How would the solution change if the GM were different?
4. How can you use these means to estimate the range of two observations?
5. What is the relationship between the AM, GM, and HM in a set of more than two observations?

**Tip:** The Harmonic Mean is most useful when averaging rates or ratios, particularly when the quantities have the same units but are distributed unevenly.

Discover how to find two observations when given their Arithmetic Mean (AM) and Geometric Mean (GM). Follow a step-by-step solution involving quadratic equations and the AM-GM inequality. Suitable for Grades 9-12.

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