Problem 5 - 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. \(\pm 2 i \sqrt{2}\) $$x^{2}=-8$$

Step-by-Step Solution

Answer

Match the solution to option H: \(\pm 2i\sqrt{2}\).

Step 1: Identify the Given Equation

The given equation is \(x^2 = -8\). Our task is to determine the solution(s) for \(x\) that satisfy this equation.

Step 2: Interpret the Equation

The equation \(x^2 = -8\) can be interpreted as a quadratic equation in the form of \(x^2 + 8= 0\). This implies that no real numbers will satisfy because the square of any real number is non-negative, and it cannot equal \(-8\). Therefore, we need to look for complex solutions.

Step 3: Solve for Complex Solutions

Since \(x^2 = -8\), we can rewrite this equation as \(x^2 = -(2^2 imes 2)\), or \(x^2 = -4 imes 2\). To solve for \(x\), take the square root of both sides, noting the imaginary unit \(i=\sqrt{-1}\): \(x = \pm \, \sqrt{-8} = \pm \, \sqrt{-4 \times 2} = \pm \, 2 \sqrt{2} i\).

Step 4: Match with Column II

The solutions \(\pm 2\sqrt{2}i\) corresponds to option H, \(\pm 2i\sqrt{2}\), in Column II.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and appear in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) cannot be zero. The solutions to these equations are the values of \( x \) that make the equation true. While some quadratics have real solutions, others involve complex numbers when the discriminant \( b^2 - 4ac \) is negative.
Here's how you solve a typical quadratic:

• Factor the quadratic expression, if possible.
• Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) when factoring is not possible.
• Complete the square to solve quadratics that are not easily factorable.

Understanding how these methods work is crucial, especially as complex solutions come into play in certain scenarios, such as when dealing with negative discriminants.

Imaginary Unit

The imaginary unit \( i \) is a key concept in complex numbers. It is defined as \( i = \sqrt{-1} \), which allows us to extend the real number system to include solutions to equations like \( x^2 = -1 \).
This imaginary unit helps us manage the square roots of negative numbers. Calculations that use \( i \) might seem unusual at first but are very handy in mathematics:

• \( i^2 = -1 \): This basic property is often used to simplify powers of \( i \), where other powers can be resolved in a cycle (e.g., \( i^3 = -i \), \( i^4 = 1 \)).
• Complex numbers in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, can model and solve real-world problems that involve periodic phenomena (e.g., wave patterns).

Having a grasp of \( i \) enriches your mathematical toolkit, broadening the range of equations you can solve.

Solutions of Equations

Solutions to equations, especially quadratics, can fall into various categories:

• Real solutions occur when the equation is satisfied with real numbers. These arise if the square root of the discriminant is non-negative.
• Complex solutions involve the imaginary unit \( i \) and emerge when the equation has no real solutions. These are especially common when dealing with negative discriminants as shown in the quadratic equation \( x^2 = -8 \).

In the example \( x^2 = -8 \), we find complex solutions \( x = \pm 2\sqrt{2}i \). Recognizing when solutions are complex is vital, as it indicates when to apply complex arithmetic and concepts such as \( a + bi \) to further manipulate and understand solutions. By grasping these solution categories, you can tackle more diverse mathematical challenges efficiently.