Determining Two Positive Numbers with a Given Difference and Product: A Comprehensive Guide

When faced with a problem involving the determination of two positive numbers that meet specific conditions, such as a given difference and product, the task can be approached through a methodical process. This guide walks you through the steps using the specific problem where the difference of two positive numbers is 5 and their product is 150.

Understanding the Problem

Let's denote the two positive numbers as x and y. The problem specifies the following conditions:

The difference between the two numbers is 5: x - y 5. The product of the two numbers is 150: xy 150.

Solving the System of Equations

To solve this, we can set up a system of equations and solve it step-by-step.

Step 1: Express One Variable in Terms of the Other

First, we express one of the variables in terms of the other. Let's express x in terms of y using the first equation:

x y 5

Step 2: Substitute into the Second Equation

Next, we substitute x in the second equation (product equation) with the expression we found:

(y 5)y 150

Expanding this, we get:

y^2 5y 150

Subtracting 150 from both sides to set the equation to zero:

y^2 5y - 150 0

Step 3: Solve the Quadratic Equation

The above equation is a quadratic equation of the form ay^2 by c 0, where a 1, b 5, and c -150. To solve this, we can either factor it or use the quadratic formula. Factoring often provides a quicker solution.

For factoring, we need two numbers that multiply to -150 and add to 5. The numbers 15 and -10 satisfy these conditions:

y^2 5y - 150 (y 15)(y - 10) 0

Step 4: Find the Solutions

Setting each factor to zero gives us:

y 15 0 implies y -15, which is not a positive number. y - 10 0 implies y 10, which is a positive number.

With y 10, we substitute back into the expression for x:

x y 5 10 5 15

Conclusion

The two positive numbers that satisfy the conditions are x 15 and y 10. These numbers meet the given requirements:

The difference is 5: 15 - 10 5. The product is 150: 15 * 10 150.

Common Pitfalls and Solutions

It's important to note that the problem specifies positive numbers. Solutions involving negative values, such as 15 and -10, do not satisfy the condition for positive numbers. Therefore, when solving such problems, it's crucial to filter out negative solutions.

For completeness, let's consider the general solution for quadratic equations:

The quadratic formula is x [-b ± sqrt(b^2 - 4ac)] / (2a). Plugging in our values, we get: a 1, b 5, and c -150. x [-5 ± sqrt(5^2 - 4(1)(-150))] / (2(1)). x [-5 ± sqrt(25 600)] / 2 -5 ± sqrt(625) / 2. x [-5 ± 25] / 2. This gives us the solutions x 15 and x -10.

Again, only the positive solution is valid for this problem.