Problem 15 - Precalculus : Building Concepts and Connections

Question

Use the definition of i to solve the equation. $$x^{2}=-16$$

Step-by-Step Solution

Answer

The solutions for the equation \(x^{2} = -16\) are \(x = 4i\) and \(x = -4i\).

Step 1: Understand the primal form of the Equation

The initial equation is given as \(x^{2} = -16\). Start by acknowledging that the solution for this equation does not lie in the real number domain since squaring a real number cannot yield a negative result.

Step 2: Express -16 as a product of 16 and -1

Express -16 as a product of 16 and -1. In that case, rewrite \(x^{2} = -16\) as \(x^{2} = 16*(-1)\).

Step 3: Apply the definition of i

Recall that 'i' is defined as \(\sqrt{-1}\). Therefore, equating that '-1' inside the root with 'i', the equation will become \(x^{2} = 16*i^{2}\) where \(i^{2}=-1\).

Step 4: Solve for x

Finally, taking the square root on both sides of the equation, we get \(x = \sqrt{16*i^{2}}\), which leads to two solutions: \(x = 4i\) and \(x = -4i\). These are the complex roots of the given equation.

Key Concepts

Imaginary Unit

When solving equations like the one given in the exercise, the concept of the imaginary unit (often denoted by \(i\)) becomes crucial. The imaginary unit is defined as \(i = \sqrt{-1}\). This definition is central because it allows us to handle square roots of negative numbers, which are not possible with real numbers alone.
When we square \(i\), we end up with:

• \(i^2 = -1\)

The significance of \(i\) lies in its ability to expand the real number system into what is known as the complex number system. Complex numbers have a real part and an imaginary part, and they are written as \(a + bi\), where \(a\) and \(b\) are real numbers. Understanding \(i\) helps us greatly when dealing with quadratic equations that don't have real solutions.

Quadratic Equations

Quadratic equations are expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The solutions to these equations can be found using various methods like factoring, completing the square, or applying the quadratic formula.
However, not all quadratic equations have real number solutions. In cases where the discriminant (\(b^2 - 4ac\)) is negative, the solutions involve complex numbers. This is because a negative discriminant indicates that the roots of the equation are not real numbers, making the imaginary unit \(i\) necessary to express these roots.
For example, in the equation \(x^2 = -16\):

• The solution requires recognizing the negative sign and expressing it using \(i\).
• This provides two solutions as complex conjugates, \(x = 4i\) and \(x = -4i\).

Using complex numbers helps resolve the equation thoroughly and reveals new insights beyond the restrictions of real numbers.

Complex Roots

Complex roots are often encountered in quadratic equations when the solutions extend beyond the real number line. A complex root arises whenever the process of solving an equation involves the square root of a negative number.
This type of root has the format \(a \pm bi\), embracing both a real part \(a\) and an imaginary part \(bi\). The given equation \(x^2 = -16\) showcases a scenario where such roots emerge:

• Recognizing \(-16\) as \(16(-1)\), allows us to use \(i\) to simplify it to \(16i^2\).
• The solutions become \(x = 4i\) and \(x = -4i\), demonstrating both positive and negative imaginary solutions.

Understanding and calculating complex roots enable mathematicians and students to gain a fuller comprehension of algebraic solutions and extend their problem-solving capabilities into the complex plane.