Finding the Numbers: A Step-by-Step Guide
Have you ever encountered a math problem that asks you to find two numbers that both multiply to a specific value and add up to another? This is a common type of problem that often appears in algebra and number theory. In this case, we're trying to find two numbers that multiply to 35 and add up to 2. Let's break down how to solve this problem.
Understanding the Problem
The problem asks us to find two numbers, let's call them 'x' and 'y', that satisfy two conditions:
- x * y = 35 (The product of the two numbers is 35)
- x + y = 2 (The sum of the two numbers is 2)
Methods for Finding the Numbers
There are a few ways to approach this problem. Here are two common methods:
1. Factoring and Trial and Error
- Step 1: Factor 35. The factors of 35 are 1, 5, 7, and 35.
- Step 2: Try different combinations of factors. See if any pair of factors adds up to 2.
- 1 + 35 = 36 (Doesn't work)
- 5 + 7 = 12 (Doesn't work)
Since none of the factor pairs add up to 2, this method doesn't lead to a solution. This indicates that there are no real integers that satisfy the conditions.
2. Using Equations
- Step 1: Solve one equation for one variable. Let's solve the second equation for 'x':
- x + y = 2
- x = 2 - y
- Step 2: Substitute the expression for 'x' into the first equation.
- (2 - y) * y = 35
- Step 3: Simplify and solve for 'y'.
- 2y - y² = 35
- y² - 2y + 35 = 0
- Step 4: Solve the quadratic equation. This equation doesn't factor easily. We can use the quadratic formula:
- y = [-b ± √(b² - 4ac)] / 2a
- Where a = 1, b = -2, and c = 35
- y = [2 ± √((-2)² - 4 * 1 * 35)] / 2 * 1
- y = [2 ± √(-136)] / 2
- y = [2 ± 2√34i] / 2
- y = 1 ± √34i
This shows that the solutions for 'y' are complex numbers (involving the imaginary unit 'i').
- Step 5: Substitute the values of 'y' back into the equation for 'x'.
- x = 2 - (1 ± √34i)
- x = 1 ∓ √34i
Therefore, the solutions are:
- x = 1 + √34i and y = 1 - √34i
- x = 1 - √34i and y = 1 + √34i
Conclusion
While there are no real integers that multiply to 35 and add up to 2, we found two complex number solutions. This demonstrates how equations can be used to find solutions even in cases where traditional factoring methods don't work. Remember that complex numbers are an important part of mathematics and can be used to solve problems that have no real number solutions.