Problem 100 - Precalculus with Limits

Question

Solving an Equation In Exercises \(97-104,\) use the formula on page 446 to find all solutions of the equation and represent the solutions graphically. $$x^{3}-27=0$$

Step-by-Step Solution

Answer

The solution for the equation \(x^{3}-27=0\) is \(x = 3\).

Step 1: Identify a and b values

In the equation \(x^{3}-27=0\), it can be transformed to the form \(x^{3}-3^{3}=0\). Here, \(a = x\) and \(b = 3\).

Step 2: Apply the formula for \(a^3 - b^3\)

We know that \(a^3 - b^3 = (a - b)\). So, the equation can be written as \(x - 3 = 0\).

Step 3: Solve for x

Solving the equation \(x - 3 = 0\), we get \(x = 3\).

Step 4: Graphical Representation

On the graph, mark \(x = 3\) on the x-axis. This is the graphical representation of the solution.

Key Concepts

Factorization

When solving cubic equations like \(x^{3} - 27 = 0\), understanding factorization is crucial. In this equation, the expression can be factored using the formula for the difference of cubes. The general formula for \(a^3 - b^3\) is given by:

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

To factor \(x^3 - 27\), identify \(a\) and \(b\) such that \(a = x\) and \(b = 3\), because \(27 = 3^3\). Applying the formula, we rewrite the expression as:

• \((x - 3)(x^2 + 3x + 9) = 0\)

Recognizing this pattern aids in breaking down the problem into more manageable terms. Notice that the equation \(x - 3 = 0\) gives a real solution, while \(x^2 + 3x + 9 = 0\) would be solved further for any complex solutions.

Graphical Representation

Representing the solutions of a cubic equation graphically can offer more insight into the problem. When graphing \(x^3 - 27 = 0\), the goal is to pinpoint where the curve intersects the x-axis. Consider the two factors obtained:

• \(x - 3 = 0\)
• \(x^2 + 3x + 9 = 0\)

The factor \(x - 3 = 0\) implies a real root of \(x = 3\). This is shown on the graph by a point at \(x = 3\) on the x-axis.
The second factor, \(x^2 + 3x + 9 = 0\), does not yield real roots but instead complex ones. This particular piece affects the shape of the curve without intersecting the x-axis. Understanding these intersections helps identify the solutions and shapes the cubic curve on a graph.

Polynomial Solutions

Finding solutions to polynomials, particularly cubic equations like \(x^3 - 27 = 0\), involves several steps. This specific equation factors into linear and quadratic polynomials:

• Linear: \(x - 3 = 0\)
• Quadratic: \(x^2 + 3x + 9 = 0\)

Solving the linear equation is straightforward, giving the real solution \(x = 3\).
The quadratic equation \(x^2 + 3x + 9 = 0\), on the other hand, requires using the quadratic formula:

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

In this equation, \(a = 1\), \(b = 3\), and \(c = 9\), leading to:

• \(x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1}\)
• Since the discriminant is negative, \(b^2 - 4ac = -27\), the solutions are complex: \(x = \frac{-3 \pm i\sqrt{27}}{2}\)

Understanding this process helps in tackling more complex polynomial equations by breaking them down into simpler factors and using the appropriate method for each.