Problem 32 - 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. $$\begin{aligned} &2 x^{3}+2 x^{2}+12 x=0\\\ &[-10,10] \text { by }[-20,20] \end{aligned}$$

Step-by-Step Solution

Answer

Solutions: \(x = 0, \frac{-1 \pm i\sqrt{23}}{2}\). Real solution verified graphically at \(x = 0\).

Step 1: Factor the Equation

First, factor out the greatest common factor from the equation: \(2x^3 + 2x^2 + 12x = 0\). The greatest factor is \(2x\). So, \(2x(x^2 + x + 6) = 0\).

Step 2: Solve for Real Solutions

Set each factor to zero and solve for the variable \(x\). This gives us two parts: 1. \(2x = 0\) which results in \(x = 0\),2. \(x^2 + x + 6 = 0\) which is solved using the quadratic formula.

Step 3: Apply the Quadratic Formula

Use the quadratic formula to solve \(x^2 + x + 6 = 0\) where \(a = 1, b = 1, c = 6\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]\[x = \frac{-1 \pm \sqrt{1 - 24}}{2}\]\[x = \frac{-1 \pm \sqrt{-23}}{2}\]Therefore, the solutions are \(x = \frac{-1 + i\sqrt{23}}{2}\) and \(x = \frac{-1 - i\sqrt{23}}{2}\).

Step 4: Write the Solution Set

The solutions to the equation \(2x^3 + 2x^2 + 12x = 0\) are \(x = 0\), \(x = \frac{-1 + i\sqrt{23}}{2}\), and \(x = \frac{-1 - i\sqrt{23}}{2}\).

Step 5: Graphical Analysis

Graph the equation \(y_1 = 2x^3 + 2x^2 + 12x\) in the viewing window \([-10, 10]\) by \([-20, 20]\). Using a calculator's graphing function, identify that the curve crosses the x-axis at the real value \(x = 0\). This supports the real solution.

Key Concepts

Factoring Polynomials

Factoring is a method used to break down a polynomial into simpler components, called factors, that when multiplied together give back the original polynomial. The first step in solving polynomial equations is often to factor them. This makes equations easier to solve.

Let's take the given equation: \(2x^3 + 2x^2 + 12x = 0\) as an example. Start by identifying the greatest common factor (GCF) among all the terms. In this equation, the GCF is \(2x\), because each term has a common factor of \(2x\).

• Divide each term by \(2x\), which is the GCF.
• This simplifies our polynomial to \(2x(x^2 + x + 6) = 0\).

By factoring out the GCF, the equation becomes easier to solve, as you end up with simpler polynomials that can each be solved individually. Solving known factors can also uncover the roots of the equation.

Quadratic Formula

When factoring is not straightforward, like in part of our example equation \(x^2 + x + 6 = 0\), we use the quadratic formula. This formula is vital for finding the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\).

The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In the example, \(a = 1\), \(b = 1\), and \(c = 6\). Plug these values into the formula:

• Calculate the discriminant: \(b^2 - 4ac = 1 - 24 = -23\).
• The negative discriminant shows that there are no real solutions, but instead, complex solutions.
• Solve for \(x\) to get \(x = \frac{-1 \pm i\sqrt{23}}{2}\).

The quadratic formula not only helps find real solutions but can easily be used to find complex solutions when the discriminant is negative.

Graphing Equations

Graphing equations is a powerful visual tool to understand the solutions of polynomials. For our original equation \(2x^3 + 2x^2 + 12x = 0\), graphing helps identify where the equation equals zero.

To graph, we need a clear viewing window that lets us see significant parts of the graph. Here, the recommended window is

• \([-10, 10]\) on the x-axis,
• \([-20, 20]\) on the y-axis.

Plotting \(y_1 = 2x^3 + 2x^2 + 12x\) will show that it crosses the x-axis at \(x = 0\), confirming the real solution. Graphing supports what algebra tells us by displaying solutions visually, which is a reliable way to verify results or identify any mistakenly overlooked solutions in complex calculations.

Real and Complex Solutions

Understanding and identifying both real and complex solutions is important when solving polynomial equations.

Real solutions are the values of \(x\) where the equation equals zero and can be seen on the graph as the points where it crosses the x-axis. Complex solutions, however, don't appear as straightforward intersections on the x-axis, as they have an imaginary component, indicated by \(i\).

• The real solution in our example is \(x = 0\), which is confirmed by graphing.
• To find complex solutions, we'd solve the part of the equation where the quadratic formula yields a negative discriminant.
• In this case, complex solutions are \(x = \frac{-1 + i\sqrt{23}}{2}\) and \(x = \frac{-1 - i\sqrt{23}}{2}\).

These solutions aren’t visible on the graph but are crucial for a complete solution set. Understanding and calculating both types of solutions ensure a thorough understanding of the behaviors of polynomial equations.