Problem 29 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation analyrically for all complex solutions, giving exact forms in your solution set. Then graph the left side of the equation as \(Y_{1}\) in the suggested viewing window and, using the capabilities of vour calculater, suppert the real solutions. $$\begin{aligned}&-3 x^{3}-x^{2}+6 x=0\\\&[-4,4] \text { by }[-10,10]\end{aligned}$$

Step-by-Step Solution

Answer

Solutions: \(x = 0, \frac{-1 + \sqrt{73}}{6}, \frac{-1 - \sqrt{73}}{6}\).

Step 1: Identify the Common Factor

The given equation is \(-3x^3 - x^2 + 6x = 0\). The first step in solving this polynomial equation is to identify and factor out the greatest common factor (GCF) from each term. Here, the common factor is \(x\). So, we factor \(x\) out: \(-3x(x^2 + \frac{1}{3}x - 2) = 0\).

Step 2: Solve the Factored Equation

Now that we have factored the equation, \(-3x(x^2 + \frac{1}{3}x - 2) = 0\), set each factor equal to zero. We have two factors to solve: \(-3x = 0\) and \(x^2 + \frac{1}{3} x - 2 = 0\). The solution to \(-3x = 0\) is \(x = 0\).

Step 3: Solve the Quadratic Equation

Next, solve the quadratic \(x^2 + \frac{1}{3}x - 2 = 0\) using the quadratic formula. The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = \frac{1}{3}\), \(c = -2\). Substitute these into the quadratic formula to get the solutions. Calculate the discriminant \(b^2 - 4ac = \frac{1}{9} + 8 = \frac{73}{9}\).

Step 4: Apply the Quadratic Formula

Substitute back into the quadratic formula: \(x = \frac{-\frac{1}{3} \pm \sqrt{\frac{73}{9}}}{2}\). Simplifying, we find \(x = \frac{-1 \pm \sqrt{73}}{6}\). Thus, the quadratic solutions are \(x = \frac{-1 + \sqrt{73}}{6}\) and \(x = \frac{-1 - \sqrt{73}}{6}\).

Step 5: Compile the Solution Set

Combine all the solutions: \(x = 0\), \(x = \frac{-1 + \sqrt{73}}{6}\), and \(x = \frac{-1 - \sqrt{73}}{6}\). So, the complete solution set of the original equation is \(\{0, \frac{-1 + \sqrt{73}}{6}, \frac{-1 - \sqrt{73}}{6}\}\).

Step 6: Graph the Left Side of the Equation

Graph \(Y_1 = -3x^3 - x^2 + 6x\) in the viewing window \([-4,4]\) by \([-10,10]\). Observe the intersection points with the x-axis to verify the real solutions. The graph should intersect the x-axis at \(x = 0\) confirming the real root.

Key Concepts

Polynomial Equations

A polynomial equation is an equation involving a polynomial expression, which typically includes variables raised to positive integer exponents. These equations can range from simple linear forms to complex higher-degree forms. For example, in our exercise, we encounter a cubic polynomial equation:

• The general form is \(-3x^3 - x^2 + 6x = 0\) where the highest power of \(x\) is 3.
• Cubic polynomials can have up to three solutions, which might be real or complex.

Solving polynomial equations often requires techniques such as factoring, using the quadratic formula, or applying synthetic division for higher degrees. It's important to find a common factor to simplify the equation, as seen with factoring out \(x\) in the given problem.
Understanding the structure of polynomial equations is crucial for finding the roots, which are the values of \(x\) that satisfy the equation.

Quadratic Formula

The quadratic formula is a solution method for quadratic equations, which are polynomials of degree 2. The formula provides the roots of the equation in terms of its coefficients:

• The standard quadratic equation is \(ax^2 + bx + c = 0\).
• The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

For our exercise, we derived the quadratic part as \(x^2 + \frac{1}{3}x - 2 = 0\). Using the quadratic formula:

• Coefficients \(a = 1\), \(b = \frac{1}{3}\), and \(c = -2\) are substituted into the formula.
• Calculate the discriminant \(b^2 - 4ac\) which determines the nature of the roots.

In this solution, the discriminant is \(\frac{73}{9}\), which is positive, indicating two distinct real roots. Understanding and applying the quadratic formula is essential for solving quadratic equations, especially when factoring is challenging.

Graphing Equations

Graphing equations visually represents the solutions of an equation. For polynomial equations, graphing helps identify where the function intersects the x-axis, corresponding to the real roots of the equation. In our problem, you graph:

• The function \(Y_1 = -3x^3 - x^2 + 6x\).
• A standard viewing window of \([-4,4] \text{ by } [-10,10]\) sets the frame for analysis.

When graphing, points where the curve crosses the x-axis indicate the real solutions. In this exercise, the graph intersects at \(x = 0\), verifying one of our solutions.
Graphing provides an intuitive sense of the behavior of polynomial functions and the nature of their solutions, making it a valuable tool alongside algebraic approaches.

Factoring Polynomials

Factoring is a method to simplify polynomial equations by expressing them as a product of simpler polynomials. It is a crucial step in finding the roots or solutions of the equation. In the given problem, factoring is executed as follows:

• Identify the greatest common factor (GCF) in the equation: \(-3x^3 - x^2 + 6x = 0\).
• Factor out the GCF, \(x\), resulting in \(-3x(x^2 + \frac{1}{3}x - 2) = 0\).

Once factored, each component can be individually set to zero, simplifying the process to find the roots. The linear factor \(-3x = 0\) directly yields \(x = 0\). The quadratic factor is solved using the quadratic formula.
Factoring reduces complexity and is often the first tool applied in solving polynomial equations. Mastering this technique opens doors to efficiently handling more complex algebraic expressions.