Problem 3 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ x^{6}-81 x^{2}=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify Common Factor

Examine the equation \( x^{6} - 81x^{2} = 0 \) and notice that \( x^2 \) is a common factor. Factor out \( x^{2} \) from the equation: \( x^{2}(x^{4} - 81) = 0 \).

Step 2: Solve for x in the Factored Equation

Set each factor equal to zero and solve for \( x \). First, set \( x^{2} = 0 \), which gives the solution \( x = 0 \).

Step 3: Solve the Polynomial Equation

Next, solve \( x^{4} - 81 = 0 \). Add 81 to both sides: \( x^{4} = 81 \).

Step 4: Simplify and Solve for x

Take the fourth root of both sides: \( x = \sqrt[4]{81} \). Simplify to find \( x = \pm 3 \), since \( (3^4) = 81 \) and \( ((-3)^4) = 81 \).

Step 5: Collect all Solutions

Combine the solutions from previous steps: \( x = 0, \pm 3 \). The real solutions are \( x = 0, 3, -3 \).

Key Concepts

Polynomial Equations

Polynomial equations are equations that involve terms with variables raised to whole number exponents. They can take many forms and degrees, but they always include a polynomial on one or both sides of the equation. For example, the exercise presents a polynomial equation of the sixth degree:

• The term with the highest exponent, namely, \( x^6 \), dictates that this is a sixth-degree polynomial equation.
• The equation is set equal to zero, which is a common characteristic of polynomial equations when searching for their roots.

Understanding polynomial equations is crucial in algebra because they allow us to model a wide range of real-world phenomena. The goal is often to find the values of the variable that make the equation true, known as roots or solutions.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler polynomials that can be multiplied to equal the original polynomial. This is an essential skill in algebra, especially when solving equations. In the exercise, the polynomial \( x^6 - 81x^2 = 0 \) can be factored by identifying common factors:

• We notice that each term contains \( x^2 \), which can be factored out, allowing us to rewrite the equation as \( x^2(x^4 - 81) = 0 \).
• This simplification reduces the complexity of solving the equation, breaking it down into manageable parts.

Factoring is useful because it allows us to apply the Zero Product Property, which states that if a product of factors equals zero, then at least one of the factors must be zero. This is a key step in solving polynomial equations.

Solving for Roots

To solve for roots means to find the values of the variable that satisfy the equation—essentially, the solutions. In the factored equation \( x^2(x^4 - 81) = 0 \), finding the roots involves setting each factor to zero:

• \( x^2 = 0 \) yields the root \( x = 0 \).
• \( x^4 - 81 = 0 \) can be rewritten as \( x^4 = 81 \).

The next step involves solving \( x^4 = 81 \) by taking the fourth root of 81. We find that \( x = \pm 3 \) as both 3 and -3 raised to the fourth power equal 81. Solving for roots can also often involve complex numbers, but in this exercise, we are focused on the real roots.

Real Solutions

Real solutions are the solutions of an equation that can be represented on the real number line. They exclude complex or imaginary numbers. In our exercise, after solving the polynomial equation, we identified the real solutions:

• Applying our factoring and solving techniques, we found \( x = 0, 3, -3 \).
• These solutions are real because they are numbers without an imaginary part and can be plotted on a standard number line.

Real solutions are significant because they often correspond to tangible, observable outcomes in real-world scenarios. For instance, if the polynomial equation models a physical system, the real solutions might represent measurable quantities like length, time, or temperature. Therefore, understanding real solutions helps in predicting real world behaviors accurately.