Problem 78 - Algebra for College Students

Question

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Missing numbers. Find two positive real numbers that differ by 2 and have a product of \(10 .\)

Step-by-Step Solution

Answer

The positive real numbers are \(-1 + \sqrt{11}\) and \(1 + \sqrt{11}\), approximately 2.3 and 4.3.

Step 1: Define Variables

Let the two positive real numbers be denoted as x and y. Given that they differ by 2, we can express this as: \(y = x + 2\)

Step 2: Set Up the Equation

Given that the product of the two numbers is 10, we can write: \(x(x + 2) = 10\)

Step 3: Form a Quadratic Equation

Expand and rearrange the equation to form a standard quadratic equation: \(x^2 + 2x - 10 = 0\)

Step 4: Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = -10\):\(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-10)}}{2(1)}\) \(x = \frac{-2 \pm \sqrt{4 + 40}}{2}\)\(x = \frac{-2 \pm \sqrt{44}}{2}\)\(x = \frac{-2 \pm 2\sqrt{11}}{2}\)\(x = -1 \pm \sqrt{11}\)

Step 5: Identify the Positive Solutions

Since the numbers must be positive, ignore the negative solutions. Thus, the first number is:\(x_1 = -1 + \sqrt{11}\).Calculate y using \(y = x + 2\):\(y_1 = -1 + \sqrt{11} + 2 = 1 + \sqrt{11}\)

Step 6: Approximate the Solutions

To the nearest tenth, find the approximate values of the solutions:\(\sqrt{11} \approx 3.3\), so\(x_1 \approx -1 + 3.3 = 2.3\)\(y_1 \approx 1 + 3.3 = 4.3\)

Key Concepts

quadratic formula

The quadratic formula is a powerful tool to solve any quadratic equation of the form nr{ax^2 + bx + c = 0}, where r{a}, r{b}, and r{c} are constants. It is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). To use the formula, follow these steps: Identify values for r{a}, r{b}, and r{c}. Then, substitute these values into the quadratic formula. The result of r{b^2 - 4ac} (called the discriminant) tells us the nature of the roots:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is one real solution (a repeated root).
• If it is negative, there are no real solutions, only complex ones.

For example, in the provided exercise, we had \(x^2 + 2x - 10 = 0\), giving r{a = 1}, r{b = 2} and r{c = -10}. Plugging these values into the quadratic formula, we found the exact solutions.

approximate solutions

Sometimes, the exact solutions from the quadratic formula involve irrational numbers (numbers that cannot be expressed as fractions). In such cases, we can find approximate solutions for practical use. Irrational numbers often result from taking the square root of a non-perfect square.
For instance, r{\sqrt{11}} is approximately r{3.3}. In the exercise, since the exact solutions were \(x_1 = -1 + \sqrt{11}\) and \(y_1 = 1 + \sqrt{11}\), we approximated these values for practical purposes to: r{x_1 \approx 2.3} and \(y_1 \approx 4.3\).
Approximations help simplify the results to a level where they can be applied in real-life contexts like measurements.

real numbers

Real numbers consist of all the numbers that can be found on the number line, including both rational and irrational numbers. They include integers, fractions, and irrational numbers like r{\sqrt{2}}. Real numbers can be positive, negative, or zero.
In the given exercise, we were asked to find two positive real numbers. To ensure solutions are within the realm of real numbers, after using the quadratic formula, we discarded negative solutions as they wouldn't make sense in context. For example, \(x_1 = -1 \pm \sqrt{11}\) only the positive part, \(-1 + \sqrt{11}\), was considered since we need two positive numbers.
Understanding real numbers is crucial in many math problems, especially when differentiating between feasible (real) solutions and infeasible (complex) ones.