Let's denote the two numbers as $x$ and $y$.

Step 1: Set up the equations

We know two things:

1. The sum of the numbers is 4: $x + y = 4$

2. The sum of their reciprocals is 8: $\frac{1}{x} + \frac{1}{y} = 8$

Step 2: Express $y$ in terms of $x$

From the first equation, we can solve for $y$: $y = 4 - x$

Step 3: Substitute $y$ into the reciprocal equation

Substitute $y = 4 - x$ into the reciprocal equation: $\frac{1}{x} + \frac{1}{4 - x} = 8$

Step 4: Simplify the equation

Combine the terms over a common denominator: $\frac{(4 - x) + x}{x(4 - x)} = 8$

Simplify the numerator: $\frac{4}{x(4 - x)} = 8$

Step 5: Cross-multiply to solve for $x$

$4 = 8x(4 - x)$

$4 = 32x - 8x^2$

$8x^2 - 32x + 4 = 0$

Step 6: Solve the quadratic equation

Divide the equation by 4 to simplify: $2x^2 - 8x + 1 = 0$

Using the quadratic formula, where $a = 2$, $b = -8$, and $c = 1$: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}$

$x = \frac{8 \pm \sqrt{64 - 8}}{4}$

$x = \frac{8 \pm \sqrt{56}}{4}$

$x = \frac{8 \pm 2\sqrt{14}}{4}$

$x = \frac{4 \pm \sqrt{14}}{2}$

Thus, the two possible values for $x$ are: $x = 2 + \frac{\sqrt{14}}{2} \quad \text{and} \quad x = 2 - \frac{\sqrt{14}}{2}$

Step 7: Find the corresponding values of $y$

For $x = 2 + \frac{\sqrt{14}}{2}$: $y = 4 - \left(2 + \frac{\sqrt{14}}{2}\right) = 2 - \frac{\sqrt{14}}{2}$

For $x = 2 - \frac{\sqrt{14}}{2}$: $y = 4 - \left(2 - \frac{\sqrt{14}}{2}\right) = 2 + \frac{\sqrt{14}}{2}$

Final Answer

The two numbers are: $2 + \frac{\sqrt{14}}{2} \quad \text{and} \quad 2 - \frac{\sqrt{14}}{2}$

Do you want more details or have any questions?

Related Questions:

1. How do you find the sum of reciprocals in other problems?
2. What are the steps to solve a quadratic equation?
3. How can you verify the solutions of this problem?
4. What is the significance of the sum of reciprocals in number theory?
5. Can you solve this problem using graphical methods?

Tip: When solving equations involving reciprocals, it’s often helpful to use common denominators to combine terms.