what numbers add to and multiply to

3 min read 27-12-2024

Finding two numbers that satisfy both a sum and a product is a common mathematical problem. This guide provides a comprehensive approach to solving such problems, from simple cases to more complex scenarios. We'll explore various methods and provide practical examples to solidify your understanding.

Understanding the Problem

The core question is: given a sum (S) and a product (P), find two numbers (x and y) such that:

x + y = S (The sum of the two numbers equals S)
x * y = P (The product of the two numbers equals P)

This seemingly simple problem has several approaches, each with its strengths and weaknesses. Let's explore them.

Method 1: Trial and Error (For Simple Cases)

For small values of S and P, trial and error can be effective. Simply list pairs of numbers that multiply to P and check if their sum is S.

Example: Find two numbers that add up to 5 and multiply to 6.

Let's try some pairs:

1 and 6: 1 + 6 = 7 (Incorrect sum)
2 and 3: 2 + 3 = 5 (Correct sum!), and 2 * 3 = 6 (Correct product!)

Therefore, the two numbers are 2 and 3.

This method is limited to simple problems. It becomes impractical for larger numbers or when dealing with negative numbers or fractions.

Method 2: Using a System of Equations

A more robust approach involves solving a system of two equations with two variables:

x + y = S
x * y = P

We can solve for one variable in terms of the other and substitute:

From equation 1, we get y = S - x. Substituting into equation 2:

x * (S - x) = P

This simplifies to a quadratic equation:

x² - Sx + P = 0

This quadratic equation can be solved using the quadratic formula:

x = [S ± √(S² - 4P)] / 2

Once you find x, you can easily find y using y = S - x.

Example: Find two numbers that add up to 7 and multiply to 12.

Here, S = 7 and P = 12. Plugging into the quadratic formula:

x = [7 ± √(7² - 4 * 12)] / 2 = [7 ± √1] / 2

This gives us two solutions: x = 3 and x = 4.

If x = 3, then y = 7 - 3 = 4. If x = 4, then y = 7 - 4 = 3. Both solutions yield the same pair of numbers: 3 and 4.

Method 3: Factoring (When Applicable)

If the quadratic equation obtained in Method 2 can be easily factored, this provides a quicker solution.

Example: Find two numbers that add up to 8 and multiply to 15.

The quadratic equation is x² - 8x + 15 = 0. This factors to (x - 3)(x - 5) = 0.

Therefore, x = 3 or x = 5. If x = 3, y = 5, and if x = 5, y = 3. The numbers are 3 and 5.

Handling Negative Numbers and Fractions

The methods above work equally well with negative numbers and fractions. Just be careful with the signs when performing calculations.

Example (Negative Numbers): Find two numbers that add up to -5 and multiply to 6.

Using the quadratic formula (S = -5, P = 6):

x = [-5 ± √((-5)² - 4 * 6)] / 2 = [-5 ± √1] / 2

This gives x = -2 and x = -3. Therefore, the numbers are -2 and -3.

Example (Fractions): Find two numbers that add up to 5/2 and multiply to 1.

Using the quadratic formula (S = 5/2, P = 1):

x = [5/2 ± √((5/2)² - 4 * 1)] / 2 = [5/2 ± √(9/4)] / 2 = [5/2 ± 3/2] / 2

This gives x = 2 and x = 1/2. Therefore, the numbers are 2 and 1/2.

Conclusion

Finding two numbers that satisfy a given sum and product can be achieved through several methods. Trial and error is suitable for simple cases, while solving a system of equations or factoring the resulting quadratic equation are more general approaches that handle negative numbers and fractions effectively. Remember to always check your solutions by verifying that they satisfy both the sum and the product conditions.