Problem 7 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ x^{6}-81 x^{2}=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 3\), and \(x = -3\).

Step 1: Factor Out the Common Term

The given equation is \(x^{6} - 81x^{2} = 0\). Notice that both terms share a common factor of \(x^2\). We can factor it out from the equation: \(x^{2}(x^{4} - 81) = 0\).

Step 2: Apply the Zero Product Property

The Zero Product Property states that if a product of factors equals zero, then at least one of the factors must be zero. This gives us two equations: \(x^{2} = 0\) and \(x^{4} - 81 = 0\).

Step 3: Solve \(x^2 = 0\)

Solving \(x^2 = 0\), we get \(x = 0\).

Step 4: Solve \(x^4 - 81 = 0\)

We can rewrite \(x^4 - 81 = 0\) as \(x^4 = 81\).

Step 5: Simplify and Solve \(x^4 = 81\)

Take the fourth root of both sides: \(x = \sqrt[4]{81}\). Since \(81 = 3^4\), we have \(x = 3\) or \(x = -3\).

Step 6: List All Real Solutions

The real solutions to the equation are \(x = 0, x = 3,\) and \(x = -3\).

Key Concepts

Zero Product Property

The Zero Product Property is a fundamental concept in algebra. It states that if a product of two or more factors is zero, then at least one of the factors must be zero. This property is incredibly useful for solving polynomial equations, such as the one in our exercise.
Think of it like this: if you multiply several numbers together and get zero, at least one of those numbers had to be zero.
This means if you have an equation where two things multiply to zero, like

• \( ab = 0 \)

then either \( a = 0 \) or \( b = 0 \) or both.
This simple rule allows us to break down complex equations into simpler parts we can solve individually.

• In our exercise, after factoring the expression to \( x^2(x^4 - 81) = 0 \), we use the Zero Product Property to set each factor equal to zero: \( x^2 = 0 \) and \( x^4 - 81 = 0 \).

This first simple step makes solving such polynomial equations much more approachable.

Factoring Polynomials

Factoring polynomials is like breaking down a complex number into smaller, multiply-able components. In our context, we first identify common factors to simplify the equation.
The initial polynomial equation from the exercise is \( x^{6} - 81x^{2} = 0 \), and the common factor here is \( x^2 \).
By factoring it out, the polynomial reduces to \( x^2(x^4 - 81) = 0 \).

• Finding this common factor is crucial as it simplifies the equation, making it easier to apply the Zero Product Property.

Once the equation is factored, it can be solved more straightforwardly by addressing each factor individually.
Factoring can involve looking for common variables, recognizing patterns like difference of squares, or pulling out coefficients. For example, \( x^4 - 81 \) here is a difference of squares since \( 81 = 3^4 \). Therefore, it can be seen as a subtraction of squares providing meaningful steps in further solving the problem.

Solving Higher Degree Equations

Higher degree equations, like the sixth-degree equation in our example, can be daunting at first. But using systematic methods, such as factoring and the Zero Product Property, they become manageable.
Once you have reduced the problem using factoring, you deal with manageable polynomial degrees.

• The simplified equation \( x^2(x^4 - 81) = 0 \) provides polynomial expressions that are far simpler to handle individually.

Consider the steps to solve higher degree equations:
1. *Factoring*: Start by finding common factors or patterns to break down the complex equation.

• In our case, this meant factoring \( x^2 \) out.

2. *Applying Zero Product Property*: Split the equation into simpler parts that equal zero.

• This results in solving easier equations like \( x^2 = 0 \) and \( x^4 - 81 = 0 \).

3. *Solving Subsequently Simpler Equations*: Address each factor as a separate, smaller problem.

• For example, solving \( x^2 = 0 \) gives \( x = 0 \).
• Solving \( x^4 = 81 \) gives \( x = 3 \) and \( x = -3 \) by recognizing it as a simple root extraction problem.

By breaking down these steps for higher degree equations, students can tackle seemingly complex problems with ease.