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Magazine Chapter 1: Problem 8
\(5-60\) Find all real solutions of the equation. $$ x^{5}-16 x=0 $$
Short Answer
Expert verified
Real solutions are \( x = 0, 2, -2 \).
Step by step solution
01
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 \( x^4 - 16 = 0 \).
02
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.
03
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: \( (x^2 - 4)(x^2 + 4) = 0 \).
04
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
Real Solutions
When solving equations, especially polynomials, we often seek real solutions 鈥� 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.
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.
