Problem 77 - Algebra for College Students

Question

Find all real and imaginary solutions to each equation. $$x^{3}+8=0$$

Step-by-Step Solution

Answer

The solutions are: -2, 1 + i√3, 1 - i√3

Step 1: Identify the equation

The given equation is a cubic equation: \(x^{3} + 8 = 0\)

Step 2: Rewrite the equation in standard form

First, isolate the cubic term by subtracting 8 from both sides: \(x^{3} = -8\)

Step 3: Factor the equation

Notice that \( -8 \) is a perfect cube, since \( -8 = (-2)^3 \). Factor using the difference of cubes formula \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\): \(x^3 + 8 = x^3 + (2^3) = (x + 2)(x^2 - 2x + 4) = 0\)

Step 4: Solve for the roots

Set each factor equal to zero and solve: \(x + 2 = 0\) gives the solution \(x = -2\); \(x^2 - 2x + 4 = 0\) can be solved using the quadratic formula. The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For the equation \(x^2 - 2x + 4\), \(a = 1\), \(b = -2\), \(c = 4\): \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(4) }}{2(1)} = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = \frac{2 \pm \sqrt{12i^2}}{2} = 1 \pm i\sqrt{3}\)

Key Concepts

Difference of Cubes

A cubic equation can often be factored using special formulas.
When we see a term like \(x^3 + 8\), we recognize it as a sum of cubes.
This is because \(8\) can be written as \((2)^3\).
The general formula for the sum of cubes is:

• \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

In our specific equation, we let \(a = x\) and \(b = 2\). This allows us to factor \(x^3 + 8\) into \((x + 2)(x^2 - 2x + 4)\).
This factoring makes the equation easier to solve.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations.
A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\).
The quadratic formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In our problem, we are dealing with the quadratic equation \(x^2 - 2x + 4 = 0\). Here, \(a = 1\), \(b = -2\), and \(c = 4\).
Plugging these values into the quadratic formula, we get:

• \(x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)}\)
• \(x = \frac{2 \pm \sqrt{4 - 16}}{2}\)
• \(x = \frac{2 \pm \sqrt{-12}}{2}\)

When simplified, this gives us the solutions \(x = 1 \pm i\sqrt{3}\).

Imaginary Solutions

When solving quadratic equations, sometimes we encounter negative values under the square root.
This indicates that the solutions will be complex or imaginary numbers.
An imaginary number is a number that can be written as a real number multiplied by the imaginary unit \(i\), where \(i\) is defined as \(\sqrt{-1}\).
In our situation, we had \(\sqrt{-12}\). We rewrote this as \(\sqrt{-12} = \sqrt{12}i\).
Therefore, when we solve the quadratic equation using the quadratic formula, we get solutions involving the imaginary unit \(i\). Our solutions were:
\(x = 1 + i\sqrt{3}\) and \(x = 1 - i\sqrt{3}\).
These are imaginary solutions because they include the imaginary unit \(i\).