Problem 43 - Precalculus

Question

Find all complex solutions to the given equations. $$x^{3}+8=0$$

Step-by-Step Solution

Answer

The complex solutions are \(x = -2, 1 + i\sqrt{3}, 1 - i\sqrt{3}\).

Step 1: Rearrange the Equation

The original equation given is \(x^3 + 8 = 0\). We start by rearranging it to isolate \(x^3\), which gives us the equation \(x^3 = -8\).

Step 2: Express as a Difference of Cubes

Recognize that \(-8\) can be rewritten as \(-2^3\). Thus, we have \(x^3 = (-2)^3\). This represents a difference of cubes, which can be factored using the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Here, \(a = x\) and \(b = -2\).

Step 3: Factor the Expression

Using the identity for the difference of cubes, \(x^3 - (-2)^3\) factors into \((x + 2)(x^2 - 2x + 4) = 0\).

Step 4: Solve the Linear Factor

The equation \(x + 2 = 0\) can be solved directly to find one solution: \(x = -2\).

Step 5: Solve the Quadratic Factor

Solve the quadratic equation \(x^2 - 2x + 4 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -2\), and \(c = 4\).

Step 6: Apply the Quadratic Formula

Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(1)(4) = 4 - 16 = -12\). Since the discriminant is negative, the roots are complex: \(x = \frac{2 \pm \sqrt{-12}}{2}\).

Step 7: Simplify the Complex Solutions

Simplify \(x = \frac{2 \pm \sqrt{-12}}{2}\) to get \(x = 1 \pm i\sqrt{3}\). These are the two complex solutions.

Key Concepts

Difference of Cubes

The difference of cubes is an important algebraic concept used to factor certain types of polynomial expressions. This concept simplifies the task of finding roots for specific cubic equations.

In our given problem, we deal with a cubic equation:

• Start with the equation: \( x^3 + 8 = 0 \)
• Isolate \( x^3 \) to get \( x^3 = -8 \).
• Notice that \( -8 \) can be rewritten as \( (-2)^3 \).

The identity used for factoring a difference of cubes is:\[a^3 - b^3 = (a-b)(a^2 + ab + b^2)\]In this particular context:

• \( a = x \)
• \( b = -2 \)

So, \( x^3 - (-2)^3 \) becomes \((x + 2)(x^2 - 2x + 4) = 0\). This method breaks the equation into a product of a linear and a quadratic factor, making it easier to solve.

Quadratic Formula

The quadratic formula is a tried-and-true method to find the roots of any quadratic equation. When you face a quadratic equation like \( ax^2 + bx + c = 0 \), this formula is a reliable tool.

In our specific exercise, we arrive at the quadratic equation \( x^2 - 2x + 4 = 0 \).

Take note of these steps:

• The coefficients are: \( a = 1 \), \( b = -2 \), and \( c = 4 \).
• Apply the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Putting in the values we get:

• \( x = \frac{2 \pm \sqrt{-12}}{2} \)

This solution is straightforward, but take note that a negative discriminant (discussed next) signals the presence of complex roots.

Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) reveals the nature of its roots. It is calculated as \( b^2 - 4ac \).

Possibilities when evaluating the discriminant's value:

• If positive, expect two distinct real roots.
• If zero, a single repeated real root is present.
• If negative, be ready for two complex roots.

In our scenario with \( x^2 - 2x + 4 = 0 \):

• Calculate the discriminant: \( (-2)^2 - 4(1)(4) = 4 - 16 = -12 \).
• The negative result indicates the existence of complex roots.

Understanding the discriminant is essential as it directly affects the kind of solutions you'll encounter, guiding you toward the correct methods for solving.

Complex Roots

Complex roots occur when the discriminant of a quadratic equation is negative, indicating no real number solutions. Instead, solutions feature both real and imaginary parts.

In our problem, the evaluated discriminant was -12, which means:

• We have two complex roots: \( x = 1 \pm i\sqrt{3} \)

The term \( i \) represents the principal square root of -1, a fundamental component of complex solutions.

To simplify the roots:

• Recall the simplified form: \( x = \frac{2 \pm \sqrt{-12}}{2} \).
• Recognize that \( \sqrt{-12} \) becomes \( i\sqrt{12} = i\sqrt{4 \times 3} = 2i\sqrt{3} \).
• Thus, simplifying results in: \( x = 1 \pm i\sqrt{3} \).

Understanding complex roots involves grasping imaginary numbers and recognizing their pattern in solutions, a valuable skill for progressing in higher-level algebra.