Given that the arithmetic mean (AM) and geometric mean (GM) of two positive numbers $a$ and $b$ are 10 and 8, respectively, we need to find the values of $a$ and $b$.

Step 1: Set up the equations

The arithmetic mean of $a$ and $b$ is given by:

$\text{AM} = \frac{a + b}{2} = 10$

The geometric mean of $a$ and $b$ is given by:

$\text{GM} = \sqrt{ab} = 8$

Step 2: Solve the AM equation

From the AM equation:

$\frac{a + b}{2} = 10$

Multiply both sides by 2:

$a + b = 20$

Step 3: Solve the GM equation

From the GM equation:

$\sqrt{ab} = 8$

Square both sides to eliminate the square root:

$ab = 64$

Step 4: Solve the system of equations

We now have the system of equations:

$a + b = 20 \quad \text{(1)}$ $ab = 64 \quad \text{(2)}$

To find $a$ and $b$, we solve this system.

Step 5: Use the quadratic formula

Consider the quadratic equation whose roots are $a$ and $b$:

$x^2 - (a + b)x + ab = 0$

Substituting from equations (1) and (2):

$x^2 - 20x + 64 = 0$

Step 6: Solve the quadratic equation

We can solve this using the quadratic formula:

$x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(1)(64)}}{2(1)}$

$x = \frac{20 \pm \sqrt{400 - 256}}{2}$

$x = \frac{20 \pm \sqrt{144}}{2}$

$x = \frac{20 \pm 12}{2}$

So, the two possible values for $x$ are:

$x = \frac{20 + 12}{2} = \frac{32}{2} = 16$ $x = \frac{20 - 12}{2} = \frac{8}{2} = 4$

Final Answer:

The two numbers $a$ and $b$ are 16 and 4.

Would you like further details or have any questions?

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Here are 5 related questions to further your understanding:

1. How do arithmetic and geometric means relate in inequalities?
2. Can you find the harmonic mean of the numbers 16 and 4?
3. What happens if $a$ and $b$ are negative?
4. How would you solve the problem if only the arithmetic mean was given?
5. Can you solve for $a$ and $b$ if their harmonic mean is given?

Tip: When dealing with quadratic equations, always check if the discriminant is a perfect square to ensure the solutions are real and rational.