Problem 25 - 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. $$\begin{aligned}&-2 x^{3}-x^{2}+3 x=0\\\&[-4,4] \text { by }[-10,10]\end{aligned}$$

Step-by-Step Solution

Answer

The complex solutions are \(x = 0\), \(x = -1.5\), and \(x = 1\).

Step 1: Identify the Equation

Given the equation \[-2x^3 - x^2 + 3x = 0\]we need to find all complex solutions. We will solve this polynomial equation analytically.

Step 2: Factor Out Common Terms

The equation can be factored by first factoring out the greatest common factor of all the terms. Here, \(x\) is common in all terms, so we can factor it out:\[x(-2x^2 - x + 3) = 0\]

Step 3: Solve for x from Factored Terms

Since we have a product of factors equaling zero,\[x = 0\]is one solution immediately. For the remaining quadratic factor, we now solve \[-2x^2 - x + 3 = 0\]

Step 4: Solve the Quadratic Equation

The quadratic equation can be solved using the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].In this equation, \(a = -2\), \(b = -1\), and \(c = 3\). Substituting these values in, we have:\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-2)(3)}}{2(-2)}\]\[x = \frac{1 \pm \sqrt{1 + 24}}{-4}\]\[x = \frac{1 \pm 5}{-4}\]

Step 5: Determine the Exact Solutions

Compute the two solutions from the quadratic:\[x = \frac{1 + 5}{-4} = -1.5\]\[x = \frac{1 - 5}{-4} = 1\]Thus, the three solutions to the equation are:\[x = 0, \; x = -1.5, \; x = 1\].

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions that involve a sum of powers of variables. The variable's highest power determines the degree of the polynomial. For example, the equation given in the exercise, \(-2x^3 - x^2 + 3x = 0\), is a cubic polynomial because the highest power is 3.
To solve polynomial equations, we aim to find all possible values of the variable(s) that make the equation true. These values are called the solutions. Polynomial equations can have real or complex solutions, and the number of solutions is equal to the degree of the polynomial. In our exercise, since the degree is three, we expect three solutions. These solutions are precisely the x-intercepts or roots of the polynomial when we graph it on a coordinate system.
Working with polynomial equations is fundamental in algebra because they can model various real-world situations, such as projectile motion, economics, and engineering challenges.

Factoring

Factoring is a method used to simplify polynomial equations by breaking them down into simpler expressions called factors. These factors, when multiplied together, give back the original polynomial. It's one of the first steps in finding the roots of a polynomial equation.

In the exercise, we begin solving the equation \(-2x^3 - x^2 + 3x = 0\) by factoring out the greatest common factor, which is the variable \(x\). This simplifies our equation to \(x(-2x^2 - x + 3) = 0\).

The key to factoring is recognizing when terms share a common factor or can be grouped to make the polynomial simpler. After factoring, each distinct factor that equates to zero will provide a solution to the equation. With simple factoring, we immediately find that one solution is \(x = 0\). The remaining quadratic factor, \(-2x^2 - x + 3 = 0\), then requires further solving, often through the quadratic formula.

Quadratic Formula

When polynomial equations involve a quadratic expression (degree of 2), one powerful method to find solutions is the quadratic formula. This formula is especially useful when other methods like simple factoring are not feasible.

The general form of a quadratic equation is \(ax^2 + bx + c = 0\) and the quadratic formula is given by:

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

In our exercise, the factor \(-2x^2 - x + 3 = 0\) is tackled with this formula. By substituting \(a = -2\), \(b = -1\), and \(c = 3\), it yields the solutions: \(x = -1.5\) and \(x = 1\).
The quadratic formula works by calculating the vertex of the parabola and finding where it intersects the x-axis. The term under the square root, \(b^2 - 4ac\), is known as the discriminant and it determines the nature of the roots:

• Positive: two real solutions
• Zero: one real solution
• Negative: two complex solutions

Using the quadratic formula guarantees that all solutions, real or complex, are found thoroughly.

Graphing Equations

Graphing equations is a visual approach to understanding the solutions and behavior of a function. In the context of polynomial equations, graphing can help visualize where the solutions or roots exist on the x-axis.

Graphing the polynomial \(-2x^3 - x^2 + 3x = 0\) involves plotting the function and observing where it crosses the x-axis. Each point of intersection indicates a real root of the equation. Using the viewing window \([-4,4]\) for the x-axis and \([-10,10]\) for the y-axis, we can see the function's behavior and confirm the real roots at 0, -1.5, and 1.
Graphing is not only useful for finding real solutions but also for understanding the overall shape and turning points of the polynomial function. By leveraging technology, such as graphing calculators or software, students can verify solutions found algebraically and gain insight into the function's properties.
Graphical representation is an essential complement to algebraic methods, providing clarity and enhancing comprehension of complex solutions.