Problem 76 - Algebra 2

Question

Solve each equation. $$ 81 x^{4}-1=0 $$

Step-by-Step Solution

Answer

The equation \(81x^4 - 1 = 0\) has four solutions: \(x = 1/3\), \(x = -1/3\), \(x = i/3\), and \(x = -i/3\).

Step 1: Identify the Form

First, identify that the equation \(81x^{4} - 1 = 0\) is a perfect square equation and it's in the form \(a^2 - b^2 = 0\) where here, \(a = 9x^2\) and \(b = 1\).

Step 2: Factor the Equation

Factor the equation using the difference of squares formula (which states that \(a^2 - b^2 = (a - b)(a + b)\)). The equation \(81x^4 - 1 = 0\) can be factored into \((9x^2 - 1)(9x^2 + 1) = 0\).

Step 3: Further Factorize the Equation

Each of \((9x^2 - 1)\) and \((9x^2 + 1)\) are also difference and sum of squares respectively, thus can be further factored. The equation \((9x^2 - 1)(9x^2 + 1) = 0\) can be further factored into no \(((3x)^2 - 1^2)((3x)^2 + 1^2) = 0\), which resolves to \((3x - 1)(3x + 1)(3x - i)(3x + i) = 0\). Here, \(i\) is the imaginary unit.

Step 4: Solve for x

Next, use the zero product property (which states that if a product of multiple factors equals zero, then at least one of the factors must be zero) to solve for \(x\). Set each factor equal to zero and solve for \(x\). This gives four potential solutions: \(x = 1/3\), \(x = -1/3\), \(x = i/3\), and \(x = -i/3\).

Key Concepts

Difference of Squares

The concept of the "difference of squares" is an essential tool in algebra that allows us to simplify and solve various equations. It applies to expressions of the form \(a^2 - b^2\), which factor into the product \((a-b)(a+b)\).
This special factorization occurs because when you multiply \((a-b)\) and \((a+b)\), the middle terms cancel out:

• \((a-b)(a+b) = a^2 + ab - ab - b^2 = a^2 - b^2\)

In the original exercise, the equation \(81x^4-1=0\) fits perfectly into this concept if we recognize \(81x^4\) as \((9x^2)^2\) and 1 as \(1^2\). Thus, we can express it as:

• \((9x^2 - 1)(9x^2 + 1) = 0\)

This factoring step is crucial because it simplifies the expression and sets the stage for finding solutions.

Factoring

Factoring is the process of breaking down an expression into simpler terms or factors that, when multiplied, give back the original expression.
In our context, factoring starts with recognizing patterns like the difference of squares, but it can go further, especially when dealing with complex expressions.
After the initial step of recognizing the difference of squares in the given equation, we factored \(81x^4-1\) into \((9x^2 - 1)(9x^2 + 1) = 0\).
We can continue to factor each part:

• \(9x^2 - 1 = (3x - 1)(3x + 1)\)
• \(9x^2 + 1 = (3x - i)(3x + i)\)

Both subsequent factors here involve simple binomials. Factoring like this is a valuable skill to quickly simplify equations, making them easier to solve.

Complex Numbers

Complex numbers introduce a new dimension to solving algebraic equations by including the imaginary unit, represented as \(i\), where \(i^2 = -1\).
They enable the expression of solutions that may not be possible with just real numbers.
In the solution process for \(81x^4 - 1 = 0\), after factoring \(9x^2 + 1\), we encountered \(3x - i\) and \(3x + i\) as factors.
This arises because \(x^2 + 1\) doesn't have real roots, but it does have complex roots, namely \(\pm i\).
Including complex numbers ensures we didn't lose potential solutions during the factoring process, allowing us to solve equations comprehensively and consider both real and imaginary parts.

Zero Product Property

The zero product property is a straightforward but powerful concept in algebra that states if the product of multiple factors is zero, at least one of the factors must be zero.
This rule helps solve equations by setting each factor equal to zero separately.
For the factored equation \((3x-1)(3x+1)(3x-i)(3x+i) = 0\), we apply the zero product property:

• Set \(3x - 1 = 0\), giving \(x = \frac{1}{3}\).
• Set \(3x + 1 = 0\), giving \(x = -\frac{1}{3}\).
• Set \(3x - i = 0\), giving \(x = \frac{i}{3}\).
• Set \(3x + i = 0\), giving \(x = -\frac{i}{3}\).

This process lays down each potential value for \(x\) that could solve the entire equation, both real and complex. It's an indispensable tool for solving factored polynomial equations.