Problem 40 - Algebra for College Students

Question

Discuss the possibilities for the roots to each equation. Do not solve the equation. $$x^{4}-1=0$$

Step-by-Step Solution

Answer

The roots are 1, -1, i, and -i.

Step 1: Identify the Type of Equation

The given equation is a polynomial equation of degree 4: \[x^{4} - 1 = 0\]

Step 2: Factoring the Equation

Factor the equation as a difference of squares: \[x^{4} - 1 = (x^{2})^{2} - 1^{2} = (x^{2} - 1)(x^{2} + 1) = 0\]

Step 3: Further Factoring

Factor \(x^2 - 1\) further: \[x^{2} - 1 = (x - 1)(x + 1)\]Therefore, the equation becomes: \[(x - 1)(x + 1)(x^{2} + 1) = 0\]

Step 4: Analyze Each Factor

Set each factor to zero and solve for \(x\): \[x - 1 = 0\] \[x + 1 = 0\] \[x^{2} + 1 = 0\]The first two factors give the real roots \(x = 1\) and \(x = -1\).

Step 5: Solve the Quadratic Equation

Solve for \(x\) in \(x^{2} + 1 = 0\): \[x^{2} + 1 = 0\] \[x^{2} = -1\] \[x = \text{±} i\]Thus, the remaining roots are imaginary roots: \(x = i\) and \(x = -i\).

Step 6: Conclusion

The equation \(x^{4} - 1 = 0\) has four roots in total: two real roots (\(x = 1\) and \(x = -1\)) and two imaginary roots (\(x = i\) and \(x = -i\)).

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra. It involves expressing a polynomial as a product of simpler polynomials. By breaking down complex polynomials into simpler ones, solving equations becomes more straightforward.

For instance, consider the polynomial equation \(x^{4} - 1 = 0\). We can factor this using various techniques such as recognizing patterns (like the difference of squares), grouping, or synthetic division.

In our example, \(x^{4} - 1\) is factored by noticing it as a difference of squares: \(x^{4} - 1 = (x^{2})^{2} - 1^{2} = (x^{2} - 1)(x^{2} + 1)\).

Beneath this initial factorization, \(x^{2} - 1\) can further be factored using the difference of squares again: \(x^{2} - 1 = (x - 1)(x + 1)\).
This process simplifies the polynomial into products of binomials, making it easier to find the roots.

Difference of Squares

The difference of squares is a powerful factoring technique in algebra. It applies when a polynomial is written in the form \(a^2 - b^2\).

By using the identity \(a^2 - b^2 = (a - b)(a + b)\), we can factor such polynomials quickly.
Let's take a deeper look at our original problem: \(x^{4} - 1\). We recognize \(x^{4}\) as \( (x^2)^2 \), and \(1\) as \(1^2\), making the equation fit the difference of squares form: \( (x^{2})^{2} - 1^{2} = (x^{2} - 1)(x^{2} + 1) \).

Imaginary Roots

Imaginary roots arise when solving equations involving negative values under a square root. They extend our number system beyond real numbers to include complex numbers.

For the polynomial equation \(x^{4} - 1 = 0\), after factoring, we get \(x^{2} + 1 = 0\).
To solve \(x^{2} + 1 = 0\), rearrange it to \(x^{2} = -1 \).
Taking the square root of both sides, we obtain \(x = \text{±} i \).

Here, the symbol \(i \) represents the imaginary unit, which satisfies \[ i^2 = -1 \]. So, \( \pm i \) are the imaginary roots of the equation.

Real Roots

Real roots of a polynomial are the solutions of the equation that are real numbers.

In our example, after factoring \(x^{4} - 1\), we wrote it as \( (x - 1)(x + 1)(x^{2} + 1) = 0 \).
Setting each factor to zero, we get: \( x - 1 = 0 \), \( x + 1 = 0 \), and \( x^{2} + 1 = 0 \).
The first two factors yield real roots \( x = 1 \) and \( x = -1 \).

Hence, \( x = 1 \) and \( x = -1 \) are the real roots of the polynomial equation \(x^{4} - 1 = 0 \).
These roots lie on the real number line and represent where the polynomial crosses the x-axis.