Problem 4 - A Graphical Approach to Precalculus with Limits

Question

Match each equation in Column I with its solution \((s)\) in Column \(I I\). A. \(\pm 2 i\) B. \(\pm 2 \sqrt{2}\) C. \(\pm i \sqrt{2}\) D. 2 E. \(\pm \sqrt{2} \quad\) F. \(-2\) G. \(\pm 2\) H. \(\pm 2 i \sqrt{2}\) $$x^{2}-2=0$$

Step-by-Step Solution

Answer

The solution is E. \(\pm \sqrt{2}\).

Step 1: Setup the Equation

The equation we have is \(x^2 - 2 = 0\). It means we need to find values of \(x\) that make this equation true.

Step 2: Solve the Equation

First, add 2 to both sides of the equation to isolate the \(x^2\) term: \(x^2 = 2\).

Step 3: Find the Solutions

Take the square root of both sides of the equation: \(x = \pm \sqrt{2}\). This indicates that there are two solutions: \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).

Step 4: Match Solutions to Column II

From the solutions \(x = \sqrt{2}\) and \(x = -\sqrt{2}\), the matching option in Column II is E. \(\pm \sqrt{2}\).

Key Concepts

Equation Solving

Understanding how to solve an equation involves finding the value or values of the variable that make the equation true. In our exercise, this process is applied to a quadratic equation: \(x^2 - 2 = 0\). A quadratic equation usually takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In this case, \(a = 1\), \(b = 0\), and \(c = -2\). To solve the equation, we begin by isolating the term with the variable, \(x^2\). By adding 2 to both sides, we get \(x^2 = 2\). This simplifies the problem and prepares us to find the values of \(x\). Highlighting the purpose of equation solving is crucial: we aim to discover values that satisfy the original condition of the equation. This is a foundational step in mathematics and is applicable to many real-world scenarios, such as solving problems related to areas, projectile motion, and more.

Square Roots

The square root operation is essential in solving equations like \(x^2 = 2\). It helps us to "undo" the squaring of \(x\) in the equation and find the original values that, when squared, equate to 2.When we take the square root of both sides of \(x^2 = 2\), we obtain \(x = \pm \sqrt{2}\). This shows two possible solutions: \(x = \sqrt{2}\) and \(x = -\sqrt{2}\).

• \(\sqrt{2}\) is a positive square root.
• \(-\sqrt{2}\) is a negative square root.

This dual nature of solutions occurs because both \((\sqrt{2})^2 = 2\) and \((-\sqrt{2})^2 = 2\). Understanding square roots is key to mastering not only quadratic equations but also other mathematical problems involving area calculations, distance, and geometry.

Mathematical Solutions

The solutions to an equation represent the values that satisfy the equation's conditions. For our equation, \(x^2 = 2\), the solutions were found as \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). These solutions are part of what is called the solution set and are critical in validating the problem-solving process.When matching our solution set to the provided choices, we identified solutions corresponding to option E: \(\pm \sqrt{2}\). Here’s a brief step-through:

• This involves not only finding the solutions but correctly relating them to a given list of potential solutions.
• It illustrates how solutions need to be verified against possible matches to ensure accuracy.

Solutions offer insight into problem conditions and prepare us for more complex problem-solving activities. They confirm our accuracy in solving mathematical problems and strengthen our mathematical intuition and reasoning skills.