Problem 23 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 23-30: Find the solutions of the equation. $$ x^{4}-16=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 2, x = -2, x = 2i, \text{and} x = -2i \).

Step 1: Rewrite the Equation

First, analyze the equation to see if it can be factored. The given equation is \( x^4 - 16 = 0 \). Notice that it can be rewritten as \( x^4 - 4^2 = 0 \), which is a difference of squares.

Step 2: Apply the Difference of Squares Formula

Recall the difference of squares formula: \( a^2 - b^2 = (a-b)(a+b) \). Apply this to the equation \( x^4 - 4^2 = 0 \). Let \( a = x^2 \) and \( b = 4 \), so that the equation becomes \( (x^2 - 4)(x^2 + 4) = 0 \).

Step 3: Solve Each Factor Separately

Now, solve each factor separately:1. \( x^2 - 4 = 0 \)2. \( x^2 + 4 = 0 \)

Step 4: Solve for \( x^2 - 4 = 0 \)

To solve \( x^2 - 4 = 0 \):- Add 4 to both sides: \( x^2 = 4 \).- Take the square root of both sides: \( x = \pm 2 \).

Step 5: Solve for \( x^2 + 4 = 0 \)

To solve \( x^2 + 4 = 0 \):- Subtract 4 from both sides: \( x^2 = -4 \).- Since this involves the square root of a negative number, use complex numbers: \( x = \pm 2i \).

Step 6: Compile the Solutions

Combine the solutions obtained from the factors:- The real solutions from \( x^2 - 4 = 0 \) are \( x = 2 \) and \( x = -2 \).- The complex solutions from \( x^2 + 4 = 0 \) are \( x = 2i \) and \( x = -2i \).

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions that involve a sum of powers of variables with coefficients. These equations can have different degrees and are characterized by the highest power of the variable. The given equation, \(x^4 - 16 = 0\), is a fourth-degree polynomial because the highest power of \(x\) is 4.

When solving polynomial equations, the goal is to find the values of the variable that make the equation true. These values are called the roots or solutions of the equation. In many cases, especially for higher-degree polynomials, factorization techniques are employed to break down the polynomial into simpler, solvable parts.

Complex Numbers

Complex numbers come into play when dealing with square roots of negative numbers. They are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit with the property \(i^2 = -1\). In the step-by-step solution, the equation \(x^2 + 4 = 0\) led to the necessity of complex numbers.

When we rearrange the equation to \(x^2 = -4\), and take the square root, we involve complex numbers to find that \(x = \pm 2i\). Here, \(2i\) and \(-2i\) are the complex roots of the equation, showcasing how we use complex numbers to handle roots of negative values.

Factorization

Factorization involves breaking down an equation into simpler "factors" that are easier to solve. For polynomial equations, this means expressing the polynomial as a product of its factors. In the problem \(x^4 - 16 = 0\), factorization was crucial for finding the roots.

We first noticed the difference of squares, \(x^4 - 4^2\), allowing us to apply the difference of squares formula and split it into \((x^2 - 4)(x^2 + 4) = 0\). Then, each of these quadratic factors was further solved for \(x\), helping us identify the roots.

Difference of Squares

The difference of squares is a specific factorization technique used when an expression of the form \(a^2 - b^2\) is encountered. This type of expression can be rewritten as \((a - b)(a + b)\). In our equation, \(x^4 - 16\) is a difference of squares because it can be expressed as \((x^2)^2 - 4^2\).

Applying the formula, it splits into two distinct parts, \((x^2 - 4)\) and \((x^2 + 4)\). By effectively using this method, solving complicated polynomial equations becomes more manageable, allowing us to tackle each factor separately and find the individual roots easily.