Problem 5 - A Graphical Approach to College Algebra

Question

Match the equation in Column I with its solution(s) in Column II. Do not use a calculator. (II) A. \(\pm 2 i\), B. \(\pm 2 \sqrt{2}\), C. \(\pm i \sqrt{2}\), D. 2, E. \(\pm \sqrt{2}\), F. \(-2\), G. \(\pm 2\), H. \(\pm 2 i \sqrt{2}\) (I) $$x^{2}=-8$$

Step-by-Step Solution

Answer

The solution is H, \(\pm 2 i \sqrt{2}\).

Step 1: Analyze the given equation

The equation provided in Column I is \(x^2 = -8\). This equation equates a square to a negative number, which means it involves imaginary numbers in its solutions since real numbers squared are never negative.

Step 2: Identify the number form needed

When \(x^2 = -8\), the solutions for \(x\) must be a complex number with an imaginary unit \(i\), since squaring a real number always gives a non-negative result. Recall that \(i\) is defined as \(\sqrt{-1}\).

Step 3: Solve the equation for x

Start by rewriting the equation as \(x^2 = -8\). Take the square root of both sides, knowing that \(\sqrt{-1} = i\). Thus, \(x = \pm \sqrt{-8} = \pm \sqrt{8} \cdot i\).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. The key feature of a quadratic equation is the term \(x^2\), making it a second-degree polynomial. These equations can provide up to two solutions, which might be real or complex numbers, depending on the disciminant, \(b^2 - 4ac\).
Larger discriminants lead to real and distinct solutions.
If the discriminant is zero, the quadratic has one real solution. In cases where the discriminant is negative, such as in our exercise, the solutions are complex, involving imaginary numbers.
This means that quadratic equations have broad applications, allowing them to solve problems ranging from physics to finance. Although they often emerge in simple algebra, they have intricate mathematical beauty, especially when introducing complex solutions.

Complex Solutions

Complex solutions arise when the roots of a quadratic equation are not entirely real. This happens because the discriminant, \(b^2 - 4ac\), of the equation is negative, implying that there is no real number solution. In such scenarios, imaginary numbers come into play, where \(i\) is defined as \(\sqrt{-1}\).
For example, consider the equation \(x^2 = -8\).
This equation cannot be solved with real numbers since the square of a real number is always non-negative. By introducing the imaginary unit \(i\), we can transform the problem: the solutions become complex numbers of the form \(x = \pm \sqrt{8} \cdot i\).
These complex numbers contain both real and imaginary parts, offering a broader spectrum for solving mathematical problems.

Square Roots

The concept of square roots is fundamental in solving quadratic equations, especially when complex numbers and imaginary units are involved. The square root of a number \(a\) is a value that, when multiplied by itself, gives \(a\).
For positive numbers like 4, the square roots are \(\pm 2\) because both \(2 \times 2\) and \((-2) \times (-2)\) yield 4. However, for negative numbers, the situation changes.
Consider the example \(x^2 = -8\), involving a negative number under the square root. This leads us to use imaginary numbers. To find the square root of -8, rewrite it as \(\sqrt{8} \cdot i\).
Here, \(\sqrt{8}\) is a real number, and multiplying it by \(i\) acknowledges the square root of a negative value, resulting in the imaginary number \(\pm \sqrt{8} \cdot i\).
This approach enables solving equations involving negative square roots by incorporating the imaginary unit \(i\), thus expanding the possibilities of mathematical solutions.