Problem 4 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ x^{5}-16 x=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 0, x = 2,\) and \(x = -2\).

Step 1: Factor the Equation

The given equation is \(x^5 - 16x = 0\). Start by factoring out the common factor \(x\): \[x(x^4 - 16) = 0 \]

Step 2: Apply Zero Product Property

Set each factor equal to zero using the zero product property:1. \(x = 0\)2. \(x^4 - 16 = 0\)

Step 3: Solve the Simple Factor

Solve the simple equation \(x = 0\). This solution is straightforward, so \(x = 0\) is one of the solutions.

Step 4: Solve the Quartic Equation

Now, solve \(x^4 - 16 = 0\). This equation can be rewritten as a difference of squares:\[x^4 - 16 = (x^2 - 4)(x^2 + 4) = 0\]

Step 5: Solve the First Quadratic

Solve \((x^2 - 4) = 0\) using the difference of squares:\[(x - 2)(x + 2) = 0\]This gives the solutions \(x = 2\) and \(x = -2\).

Step 6: Analyze the Second Quadratic for Real Solutions

Solve \((x^2 + 4) = 0\). Rearrange to:\[x^2 = -4\]This equation has no real solutions because a square cannot be negative in the reals.

Key Concepts

Factoring Equations

Factoring equations is a powerful technique used to simplify and solve polynomial equations. It involves breaking down a polynomial into a product of its simpler polynomial factors. For instance, if you have the equation \(x^5 - 16x = 0\), you can start by identifying the common factors in each term. Here, both terms share \(x\) as a factor. By factoring \(x\) out, the equation becomes \[x(x^4 - 16) = 0.\]

• Look for a common factor that can be taken out.
• Rewriting the equation with the factored terms makes it easier to apply other mathematical techniques.

It’s essentially like breaking down a complex problem into simpler parts, which can then be solved individually or recombined for further simplification.

Zero Product Property

The zero product property is a fundamental principle used in mathematics when working with factored equations. This property states that if the product of two or more factors equals zero, then at least one of the factors must be zero. In simple terms, if \(ab = 0\), then \(a = 0\) or \(b = 0\). For the factored equation \[x(x^4 - 16) = 0,\] using the zero product property, we can set each factor to zero:

• \(x = 0\), which is one solution.
• \(x^4 - 16 = 0\), requiring further exploration.

This method simplifies solving equations because each factor can be considered separately, allowing us to find all potential solutions. It’s an efficient way to dissect a problem into manageable pieces.

Quartic Equations

Quartic equations are polynomial equations of the fourth degree, such as \(x^4 - 16 = 0\). Solving quartic equations can be challenging, but they are often easier to handle when rewritten through factoring methods like the difference of squares. The equation \(x^4 - 16\) can be factored as:

• \((x^2 - 4)(x^2 + 4) = 0\)

Each factor represents a simpler quadratic equation, making it easier to solve. In this case, by applying the factoring again, \(x^2 - 4\) breaks into two linear factors \((x - 2)\) and \((x + 2)\). Thus, it yields real solutions easily. Quartic equations sometimes require various techniques, including recognizing patterns or using substitution, to break down the expression into manageable forms.

Difference of Squares

The difference of squares is a special factoring technique used particularly for expressions like \(a^2 - b^2\). This technique relies on the identity \(a^2 - b^2 = (a - b)(a + b)\). Knowing this helps in simplifying the quartic equation \(x^4 - 16 = 0\). The expression becomes:

• \(x^4 - 16 = (x^2 - 4)(x^2 + 4)\)

The term \(x^2 - 4\) itself is another difference of squares:

• \((x^2 - 4) = (x - 2)(x + 2)\)

This sequence of factoring simplifies complex equations by breaking them into linear terms, allowing for easier solution finding. The difference of squares is particularly useful when working with equations that initially appear complex but have underlying symmetric patterns.