Problem 26 - A Graphical Approach to College Algebra

Question

Solve each equation analytically 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 your calculator, support the real solutions. $$\begin{aligned} &-5 x^{3}+13 x^{2}+6 x=0\\\ &[-4,4] \text { by }[-2,30] \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = -0.4\), and \(x = 3\).

Step 1: Factor the equation

First, observe the polynomial equation \(-5x^3 + 13x^2 + 6x = 0\). Notice that each term has an \(x\) in it, allowing us to factor out \(x\), resulting in: \(x(-5x^2 + 13x + 6) = 0\). This gives us one solution immediately: \(x = 0\).

Step 2: Solve the quadratic equation

After factoring out the \(x\), we are left with solving the quadratic equation \(-5x^2 + 13x + 6 = 0\). We can use the quadratic formula to find its solutions: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -5\), \(b = 13\), and \(c = 6\).

Step 3: Calculate the discriminant

Calculate the discriminant \(\Delta = b^2 - 4ac\). Substitute the values to get \(\Delta = 13^2 - 4(-5)(6) = 169 + 120 = 289\). The positive discriminant indicates two distinct real roots.

Step 4: Calculate the roots using the quadratic formula

Substitute the values and the discriminant into the quadratic formula: \(x = \frac{-13 \pm \sqrt{289}}{-10}\). Simplifying this gives \(x = \frac{-13 \pm 17}{-10}\). Thus, we have \(x = \frac{4}{-10} = -0.4\) and \(x = \frac{-30}{-10} = 3\).

Step 5: List all solutions

Combining both solutions from factoring and the quadratic solution, we have three solutions for \(x\): \(x = 0\), \(x = -0.4\), and \(x = 3\). These are the solutions to the original equation.

Step 6: Graph the function to support real solutions

Using a graphing calculator, plot the function \(y_1 = -5x^3 + 13x^2 + 6x\) in the viewing window \([-4, 4]\) by \([-2, 30]\). Observe the x-axis intersections occur at the points \(x = 0\), \(x = -0.4\), and \(x = 3\), which confirms the real solutions calculated analytically.

Key Concepts

Polynomial Equation

A polynomial equation is a sum of terms, each consisting of a variable raised to an integer power, multiplied by a coefficient. These equations can have multiple solutions, including real and complex numbers. In the case of the equation from the original exercise,
\(-5x^3 + 13x^2 + 6x = 0\), it is a polynomial of degree three, indicating it can have up to three solutions.

Polynomial equations are fundamental in algebra and can be solved using various methods, depending on their degree:

• **Linear Equations**: Simple solutions obtained by rearranging the terms.
• **Quadratic Equations**: Solved using factoring methods, completing the square, or the quadratic formula.
• **Higher-degree polynomials (third degree or more)**: Often solved by factoring, synthetic division, or rational root theorem among other techniques.

In this exercise, the initial step was to factor out common terms, simplifying the third-degree polynomial to a quadratic equation, which is more manageable to solve.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which have the general form \(ax^2 + bx + c = 0\). It allows you to find the roots of any quadratic equation, regardless of their discriminant, using the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides exact solutions and helps determine the nature of these solutions (real or complex) based on the discriminant. For the given quadratic equation \(-5x^2 + 13x + 6 = 0\), setting:

• \(a = -5\)
• \(b = 13\)
• \(c = 6\)

allowed us to substitute into the quadratic formula and solve for \(x\).

The results of the quadratic formula provided the additional roots \(x = -0.4\) and \(x = 3\), alongside the factorable root \(x = 0\). These are the real solutions to the polynomial equation.

Graphing Calculator

Graphing calculators are indispensable tools in visualizing functions and confirming algebraic solutions. They allow students to plot equations and visually determine where these functions intersect the axes, highlighting solutions.

When graphing the equation \(y_1 = -5x^3 + 13x^2 + 6x\) on a graphing calculator within the suggested window \([-4, 4]\) for the x-axis and \([-2, 30]\) for the y-axis, the real intersections where the curve meets the x-axis provide a visual confirmation of the roots determined:

• \(x = 0\)
• \(x = -0.4\)
• \(x = 3\)

This plotting helps validate the algebraic process and assures that the solutions are applied correctly in real contexts. Graphing calculators are not only used for confirming solutions but also for understanding the behavior and nature of polynomial equations.

Discriminant

The discriminant is a significant part of the quadratic formula, represented by \(b^2 - 4ac\). It determines the nature of the roots of a quadratic equation:

• If the discriminant is **positive**, there are two distinct real roots.
• If zero, one real root exists (or a double root).
• If negative, the roots are complex or imaginary, lacking real solutions.

In the original exercise, the discriminant calculated from the quadratic part,
\[\Delta = 13^2 - 4(-5)(6) = 169 + 120 = 289\],
was positive, indicating two real, distinct roots could be derived from the quadratic equation \(-5x^2 + 13x + 6 = 0\).

The positive discriminant led to successful application of the quadratic formula, providing the solutions \(x = -0.4\) and \(x = 3\), validating the algebraic approach and reinforcing the visual confirmation using the graphing calculator. Understanding the role of the discriminant brings insight into the nature of solutions, signaling whether to expect real or complex numbers before solving.