Unlocking Number Mysteries Find Sum -5 Product -50 Solutions

by Scholario Team 61 views

The realm of mathematics often presents us with intriguing puzzles, challenging our problem-solving skills and pushing the boundaries of our understanding. Among these captivating conundrums are those that involve finding numbers that satisfy specific conditions, such as having a particular sum and product. In this article, we embark on a journey to unravel one such mystery: identifying the numbers whose sum is -5 and whose product is -50. This exploration will not only hone our algebraic prowess but also reveal the elegance and interconnectedness of mathematical concepts.

To embark on our quest, let's first formally define the problem at hand. We are tasked with finding two numbers, which we'll denote as x and y, that simultaneously satisfy the following conditions:

  1. x + y = -5 (The sum of the numbers is -5)
  2. x y = -50 (The product of the numbers is -50)

These two equations form a system of equations, a mathematical construct that allows us to represent multiple relationships between variables. Our goal is to find the values of x and y that make both equations true. There are several methods to solve a system of equations, including substitution, elimination, and graphical techniques. In this case, we will employ the substitution method, a powerful tool that involves expressing one variable in terms of the other.

The substitution method entails the following steps:

  1. Solve one equation for one variable: Let's choose the first equation, x + y = -5, and solve for x. Subtracting y from both sides, we get:

    x = -5 - y

  2. Substitute the expression into the other equation: Now, we substitute this expression for x into the second equation, x y = -50:

    (-5 - y) y = -50

  3. Simplify and rearrange: Expanding the left side of the equation, we get:

    -5y - y² = -50

    Adding 50 to both sides and rearranging, we obtain a quadratic equation:

    y² + 5y - 50 = 0

  4. Solve the quadratic equation: We now have a quadratic equation in the variable y. There are several ways to solve quadratic equations, including factoring, using the quadratic formula, and completing the square. In this case, factoring proves to be the most straightforward approach. We seek two numbers that multiply to -50 and add to 5. These numbers are 10 and -5. Thus, we can factor the quadratic equation as:

    (y + 10)(y - 5) = 0

    Setting each factor equal to zero, we get two possible solutions for y:

    y + 10 = 0 or y - 5 = 0

    y = -10 or y = 5

  5. Substitute back to find the other variable: We have found two possible values for y. To find the corresponding values for x, we substitute each value of y back into the equation x = -5 - y:

    • If y = -10, then x = -5 - (-10) = 5
    • If y = 5, then x = -5 - 5 = -10

We have successfully navigated the steps of the substitution method and arrived at two pairs of numbers that satisfy the given conditions:

  • x = 5, y = -10
  • x = -10, y = 5

Thus, the numbers whose sum is -5 and whose product is -50 are 5 and -10. This seemingly simple problem has led us on a journey through algebraic manipulation, quadratic equations, and the power of problem-solving strategies. This exploration underscores the beauty and interconnectedness of mathematical concepts, demonstrating how different tools and techniques can be employed to unravel even the most enigmatic of numerical mysteries.

While the substitution method has proven effective in solving this problem, it's worth noting that other approaches could also be employed. For instance, one could use the quadratic formula directly on the equation y² + 5y - 50 = 0, which would yield the same solutions for y. Additionally, one could explore graphical methods, plotting the equations x + y = -5 and x y = -50 and identifying the points of intersection, which would correspond to the solutions.

Furthermore, this problem provides an opportunity to delve into the relationship between the coefficients of a quadratic equation and its roots. In general, for a quadratic equation of the form ax² + bx + c = 0, the sum of the roots is given by -b/ a, and the product of the roots is given by c/ a. In our case, the quadratic equation is y² + 5y - 50 = 0, so the sum of the roots is -5/1 = -5, and the product of the roots is -50/1 = -50, which aligns with the original problem statement.

This exploration into finding numbers with a specific sum and product has not only provided us with a solution but also offered valuable insights into algebraic techniques, problem-solving strategies, and the interconnectedness of mathematical concepts. The problem, at first glance, might have seemed like a simple exercise, but it has proven to be a gateway to a deeper understanding of mathematical principles. By employing the substitution method, we successfully identified the numbers 5 and -10 as the solution, demonstrating the power and elegance of mathematical reasoning. This journey serves as a testament to the beauty and intrigue that lie within the realm of mathematics, encouraging us to continue exploring its mysteries and challenges.

Keywords: sum, product, numbers, substitution method, quadratic equation, factoring, roots, algebra, problem-solving, mathematics.