Solving for Numbers that Satisfy Specific Multiplicative and Additive Conditions

Let's explore the intriguing problem of finding two numbers that multiply to -36 and add up to 7. This type of problem is a classic example used in algebra to illustrate the application of quadratic equations.

Setting Up the Equations

The given conditions can be simplified into two key equations:

Equation 1: The numbers multiply to -36. Equation 2: The numbers add up to 7.

We can represent the two numbers as x and y. Therefore, our equations become:

x y -36 x y 7

Expressing One Variable in Terms of the Other

From Equation 2, we can express y in terms of x as follows:

[y 7 - x]

Substituting into the First Equation

Next, we substitute this expression for y into the first equation:

[x(7 - x) -36]

This simplifies to:

[7x - x^2 -36]

By re-arranging terms, we obtain a standard quadratic equation:

[x^2 - 7x - 36 0]

Solving the Quadratic Equation

To solve the quadratic equation, we can use the quadratic formula:

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

In our equation, the coefficients are:

[a 1, quad b -7, quad c -36]

Substituting these values into the formula, we get:

[x frac{-(-7) pm sqrt{(-7)^2 - 4 cdot 1 cdot (-36)}}{2 cdot 1} ]

First, calculate the discriminant:

[(-7)^2 - 4 cdot 1 cdot (-36) 49 144 193]

Now, substituting the discriminant back into the formula:

[x frac{7 pm sqrt{193}}{2}]

Using a calculator to approximate the values of the roots:

[x_1 approx frac{7 13.89}{2} approx 10.45]

[x_2 approx frac{7 - 13.89}{2} approx -3.45]

Finding the Corresponding Values of Y

Once we have the values of x, we can find the corresponding values of y using the expression y 7 - x:

[y_1 7 - 10.45 approx -3.45]

[y_2 7 - (-3.45) approx 10.45]

Therefore, the two numbers that satisfy the conditions are approximately 10.45 and -3.45.

Summary

To summarize, the two numbers that multiply to -36 and add up to 7 are approximately:

[10.45 quad text{and} quad -3.45]

Alternative Methods

Another approach is to substitute unknown numbers with y and z and use the same principle. The steps would be:

yz -36 y z 7 (y 7 - z) ((7 - z)z -36) 7z - (z^2) -36 (-z^2 7z 36 0) (Multiplying by -1) (z^2 - 7z - 36 0) Using the quadratic formula again, we get the values for (z).

This will lead to the same solution set as the previous method, confirming the accuracy of the solution.

No Integer Solutions

It is also important to note that if we were to look for integer solutions, none of the pairs that multiply to -36 would sum to 7. The integer pairs that multiply to -36 are: [-1, 36], [1, -36], [-2, 18], [2, -18], [-3, 12], [3, -12], [-4, 9], [4, -9], and [-6, 6]. None of these pairs sum to 7, which is why there are no integer solutions to this problem.

However, the quadratic method provides us with the exact real solutions, which are approximately 10.45 and -3.45.