Problem 53 - Algebra A Combined Function

Question

The sum of two numbers is \(25,\) and the sum of their squares is \(325 .\) Find the numbers.

Step-by-Step Solution

Answer

The numbers are 10 and 15.

Step 1: Define Variables

Let the two numbers be \(x\) and \(y\). According to the problem, the sum of these two numbers is equal to \(25\), so we can write the equation: \(x + y = 25\).

Step 2: Formulate the Second Equation

The problem states that the sum of their squares is \(325\). Thus, we can write the second equation as follows: \(x^2 + y^2 = 325\).

Step 3: Express One Variable in Terms of the Other

From the first equation \(x + y = 25\), express \(y\) in terms of \(x\): \(y = 25 - x\).

Step 4: Substitute into the Second Equation

Substitute \(y = 25 - x\) into the second equation \(x^2 + y^2 = 325\) to solve for \(x\):\[x^2 + (25 - x)^2 = 325.\]

Step 5: Simplify the Equation

Expand the equation: \(x^2 + (25 - x)^2 = x^2 + (625 - 50x + x^2) = 325.\). This simplifies to \(2x^2 - 50x + 625 = 325\).

Step 6: Rearrange and Simplify Further

Bring all terms to one side: \(2x^2 - 50x + 625 - 325 = 0\). Simplify to get: \(2x^2 - 50x + 300 = 0\).

Step 7: Solve the Quadratic Equation

Divide the entire equation by \(2\) to simplify: \(x^2 - 25x + 150 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -25\), and \(c = 150\).

Step 8: Evaluate the Quadratic Formula

Compute the discriminant: \((-25)^2 - 4 \times 1 \times 150 = 625 - 600 = 25\).Compute the roots: \[x = \frac{25 \pm \sqrt{25}}{2} = \frac{25 \pm 5}{2}\].Therefore, \(x = 15\) or \(x = 10\).

Step 9: Find Corresponding Values for the Other Variable

Substitute back to find \(y\):\(y = 25 - x\).If \(x = 15\), then \(y = 25 - 15 = 10\).If \(x = 10\), then \(y = 25 - 10 = 15\).

Step 10: Conclusion

The two numbers are \(10\) and \(15\). Either \(x = 10\) and \(y = 15\) or \(x = 15\) and \(y = 10\) satisfy both conditions of the problem.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions where the highest degree of the variable is squared (degree of 2). They come in the standard form of:

• \[ ax^2 + bx + c = 0 \]

Where \(a\), \(b\), and \(c\) are constants and \(a\) is not equal to zero. In solving problems, such as the one in the exercise, quadratics are useful because they can model situations involving the sum and square of numbers.
Quadratics can be solved by multiple methods, including:

• Factoring, which is useful when the equation can be easily broken into simple products.
• Completing the square, a method involving making a perfect square trinomial.
• Quadratic Formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our problem, after forming the quadratic equation \( x^2 - 25x + 150 = 0 \), we used the quadratic formula to find the possible values of \(x\). The quadratic formula provides an efficient way to find the roots of the equation, especially when factors are not immediately obvious.

Sum of Two Numbers

The concept of the sum of two numbers involves finding two values that add up to a given total. In algebraic terms, if you have two numbers, \(x\) and \(y\), the equation representing their sum is:

• \[ x + y = \, \text{total sum} \]

In this exercise, the sum given is \(25\), so the equation becomes \(x + y = 25\).
Determining the sum is often a first step in solving problems involving relationships between numbers. This allows for the expression of one variable in terms of the other, simplifying the system into a single equation. Such simplification is crucial for tackling more complex algebraic problems, like when variables are squared or have other added conditions. This expression would aid us in the substitution and solving of a quadratic equation as demonstrated in the original example.

Systems of Equations

Systems of equations involve having multiple equations that share variables, and the goal is to find values for these variables that satisfy all of the equations simultaneously. In our case, we had two equations:

• \[ x + y = 25 \]
• \[ x^2 + y^2 = 325 \]

To manage this system, we use a method called substitution. Starting with solving one of the equations for one variable in terms of the others (\( y = 25 - x \)), allows it to replace \( y \) in the second equation.
This converts the system into a single quadratic equation in one variable, \( x \). Once the equation is simplified and solved, the value of \(x\) can be placed back into the expression for \(y\). This yields the possible solutions for the system. Solving systems of equations in this way is a fundamental skill in algebra that enables the handling of multi-variable problems in a structured manner.