Introduction

In mathematics, a common problem involves finding two numbers when their sum and product are known. This problem is not only foundational in algebra but also appears in various applications, such as in economics, physics, and engineering. This article explores the solution to the problem where the sum of two numbers is 7 and their product is -60.

Solution Approach

Let's denote the two numbers by x and y. Given the constraints:

1. x y 7

2. xy -60

We can express y in terms of x using the first equation:

y 7 - x

Substituting y 7 - x into the product equation:

x(7 - x) -60

Step 1: Simplifying the Equation

Multiplying the equation:

7x - x^2 -60

Rearrange the equation into a standard quadratic form:

x^2 - 7x - 60 0

Step 2: Solving the Quadratic Equation

We will solve this quadratic equation using the quadratic formula:

x frac{-b pm sqrt{b^2 - 4ac}}{2a}

Here, a 1, b -7, and c -60. Substituting these values, we get:

x frac{7 pm sqrt{49 240}}{2}

x frac{7 pm sqrt{289}}{2}

x frac{7 pm 17}{2}

This gives us two possible values for x:

x frac{24}{2} 12

x frac{-10}{2} -5

Step 3: Finding the Corresponding Values for y

Using y 7 - x:

If x 12:

y 7 - 12 -5

If x -5:

y 7 - (-5) 12

Therefore, the two numbers are 12 and -5.

Alternative Solutions

Another approach to solving this problem involves the method of alpha - beta and roots of a quadratic equation. Here's a simplified version of the solution:

alpha - beta pm sqrt{alpha beta^2 - 4 alpha beta}

alpha - beta pm sqrt{49 - 240}

alpha - beta pm 17

alpha beta frac{alpha beta pm alpha - beta}{2}

alpha beta frac{-7 pm 17}{2}

alpha, beta 5, -12

Since the product is negative, the numbers must be -5 and 12. Therefore, the solution is consistent with our earlier findings.

Another Realization

A third realization involves a slightly different perspective, which can be useful for cross-validation:

1. x - y 7

2. xy 60

Solve for x in the first equation:

x y 7

Substitute into the second equation:

(y 7)y 60

y^2 7y - 60 0

Factorize the quadratic equation:

(y - 5)(y 12) 0

This gives us two possible solutions:

y 5

y -12

For y 5:

x 12

For y -12:

x -5

Thus, the numbers can be 12 and 5 or -5 and -12.

Conclusion

This problem is a classic example of using systems of equations and quadratic equations in algebra. Whether using the traditional method or a simplified alternative, the solution remains the same, highlighting the importance of these fundamental mathematical tools.