Problem 8 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ x^{5}-16 x=0 $$

Step-by-Step Solution

Answer

Real solutions are \( x = 0, 2, -2 \).

Step 1: Factor the Equation

The given equation is \( x^5 - 16x = 0 \). To start, factor out the common term \( x \) from the equation: \( x(x^4 - 16) = 0 \). This gives us two separate equations to solve: \( x = 0 \) and \( x^4 - 16 = 0 \).

Step 2: Solve the Simple Factor

To solve for \( x = 0 \), we simply have \( x = 0 \) as one solution. This is self-evident since any number multiplied by zero is zero. Therefore, \( x = 0 \) is a solution to the equation.

Step 3: Solve for Remaining Factors

Next, solve \( x^4 - 16 = 0 \) by recognizing it as a difference of squares. Rewrite it as \((x^2)^2 - 4^2 = 0\), which can be factored using the difference of squares formula: \( (x^2 - 4)(x^2 + 4) = 0 \).

Step 4: Solve Each Factor Separately

Now, solve the equation \( x^2 - 4 = 0 \). This can also be factored as \((x - 2)(x + 2) = 0 \). Solving these gives \( x = 2 \) and \( x = -2 \).For \( x^2 + 4 = 0 \), recognize that it has no real solutions because \( x^2 = -4 \) cannot happen with real numbers, as squares are non-negative.

Key Concepts

Factoring Polynomials

Factoring polynomials is a fundamental step in solving polynomial equations. In essence, it involves expressing a polynomial as a product of simpler polynomials. This is analogous to breaking down a number into its prime factors. For example, in the equation \( x^5 - 16x = 0 \), we take a strategic approach by factoring out the common term. We notice that both terms include \( x \), allowing us to factor it out, resulting in \( x(x^4 - 16) = 0 \).

• This factoring simplifies the problem into two more manageable equations.
• Factoring reduces a complex equation into simpler components that can be solved independently.
• Recognizing common factors is a crucial skill in simplifying equations.

Breaking down polynomials into their factors can simplify the process of finding solutions and often reveals the straightforward solutions we might initially overlook.

Difference of Squares

The difference of squares is a specific type of polynomial factoring that occurs in expressions of the form \( a^2 - b^2 \). This type of expression can be rewritten as \( (a - b)(a + b) \). It takes advantage of the fact that a squared term subtracted by another squared term can be broken down this way.

• In our original problem, we reached \( x^4 - 16 = 0 \), which is a difference of squares.
• We rewrite \( x^4 - 16 \) as \((x^2)^2 - 4^2 \), a recognizable difference of squares.
• The equation becomes \( (x^2 - 4)(x^2 + 4) = 0 \) through this factoring method.

Using the difference of squares technique simplifies our equation and allows us to find more solutions by breaking it down into simpler factors that are easier to solve.

Real Solutions

When solving equations, especially polynomials, we often seek real solutions — values of \( x \) that satisfy the equation and belong to the set of real numbers. In the given exercise, real solutions were derived step by step.

• From \( x^4 - 16 \), after factoring using difference of squares, we considered each factor separately.
• For \( x(x^2 - 4)(x^2 + 4) = 0 \), solving \( x = 0 \) gives us one straightforward real solution.
• Further solving \( x^2 - 4 = 0 \) yields two more real solutions: \( x = 2 \) and \( x = -2 \).
• However, \( x^2 + 4 = 0 \) provides no real solutions because square values are non-negative, and no real number squared equals a negative value.

Identifying real solutions involves recognizing which values fit within the constraints of real numbers, which excludes any roots of negative numbers within real number context.

Quadratic Equations

Quadratic equations frequently appear when factoring higher degree polynomials down to simpler terms. These are equations that can be written in the form \( ax^2 + bx + c = 0 \), and their solutions provide critical insights into more complex polynomial solutions.

• In our process, after applying the difference of squares formula, we encounter \( x^2 - 4 = 0 \), a quadratic equation.
• This can be further factored into \((x - 2)(x + 2) = 0 \), allowing us to solve for \( x = 2 \) and \( x = -2 \).
• These types of equations are typically solved by factoring, using the quadratic formula, or by completing the square.
• Recognizing quadratic structures within polynomials is key to simplifying and solving them efficiently.

Mastery of quadratic equations is foundational, as this knowledge allows students to handle more complicated equations with confidence and precision.