Problem 33 - 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. $$\begin{aligned} &x^{4}+17 x^{2}+16=0\\\ &[-4,4] \text { by }[-10,40] \end{aligned}$$

Step-by-Step Solution

Answer

The solutions are \(\{i, -i, 4i, -4i\}\), with no real solutions.

Step 1: Identify the Equation Type

The given equation is a polynomial equation of the form \(x^4 + 17x^2 + 16 = 0\). To solve for \(x\), recognize that it is a quadratic in terms of \(x^2\). This allows us to perform substitution.

Step 2: Perform Substitution

Let \(u = x^2\). Then the equation becomes \(u^2 + 17u + 16 = 0\). This new equation now can be solved like a regular quadratic equation in terms of \(u\).

Step 3: Solve the Quadratic Equation

Use the quadratic formula, \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For \(u^2 + 17u + 16 = 0\), let \(a = 1, b = 17, c = 16\). Calculate the discriminant: \(b^2 - 4ac = 17^2 - 4 \times 1 \times 16 = 225\).Substitute in the formula to find \(u\): \[ u = \frac{-17 \pm \sqrt{225}}{2} = \frac{-17 \pm 15}{2} \]This gives two solutions for \(u\): \(u_1 = -1\) and \(u_2 = -16\).

Step 4: Solve for \(x\) from \(u\)

Since \(u = x^2\), we must solve two sets of equations: 1. \(x^2 = -1\)2. \(x^2 = -16\)For \(x^2 = -1\), the solutions are \(x = i, -i\).For \(x^2 = -16\), the solutions are \(x = 4i, -4i\).

Step 5: Find the Solution Set

The complete set of solutions for the original equation \(x^4 + 17x^2 + 16 = 0\) is \(\{i, -i, 4i, -4i\}\). These are complex numbers.

Step 6: Graph and Analyze Real Part

Graph \(y_1 = x^4 + 17x^2 + 16\) in the viewing window \([-4,4]\) by \([-10,40]\). Analyze that graph for any intersections with the x-axis which validate real solutions. The graph will not cross the x-axis, indicating there are no real solutions, only complex solutions.

Key Concepts

Quadratic Substitution

To solve complex polynomial equations, we sometimes use a technique called quadratic substitution. This method is particularly useful for equations that can be transformed into a quadratic form. Take the polynomial equation given: \(x^4 + 17x^2 + 16 = 0\). Though it initially appears quite complex, notice how it can be seen as a quadratic equation where \(u = x^2\). Because of this, the equation can be rewritten into a simpler form, \(u^2 + 17u + 16 = 0\), where \(u\) represents \(x^2\).

This transformation simplifies the process as you can now solve for \(u\) using quadratic methods. By changing variables in this manner, quadratic substitution brings us back to familiar ground, allowing us to leverage tools like the quadratic formula or factoring. Once \(u\) is found, we simply substitute back to solve for \(x\). In this problem, such substitution guides us to find complex solutions for \(x^4 + 17x^2 + 16 = 0\).

Polynomial Equations

Polynomial equations can sometimes feel daunting due to their complexity and many terms. However, understanding their structure can greatly simplify the solving process. In our example, \(x^4 + 17x^2 + 16 = 0\), we have a quartic (degree 4) polynomial.

Polynomial equations like this have unique properties:

• The degree gives a sense of how many solutions to expect, in this case, four.
• Solutions can be real, complex, or a combination of both.
• Each term represents powers of \(x\), indicating the polynomial's overall behavior when graphed.

After transforming the quartic equation into a quadratic one using substitution, the complexity significantly reduces. This highlights the power of breaking down polynomial equations into more manageable parts. It helps focus on core concepts without being overwhelmed by the equation's size or degree.

Quadratic Formula

The quadratic formula is one of the most powerful tools for solving quadratic equations of the form \(ax^2 + bx + c = 0\).

In the equation resulting from our substitution, \(u^2 + 17u + 16 = 0\), we set \(a = 1\), \(b = 17\), and \(c = 16\) to apply this formula:

\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The term under the square root, \(b^2 - 4ac\), is called the discriminant. It tells us the nature of the roots:

• If it's positive, we have two distinct real roots.
• If zero, one real root (a repeated solution).
• If negative, two complex roots.

In our case, the discriminant is 225, a positive number, indicating two real roots for \(u\), specifically \(u = -1\) and \(u = -16\). Yet, because \(u = x^2\), and these solutions are negative, it confirms that our final answers will be complex numbers for \(x\). Thus, \(x = \pm i\) and \(x = \pm 4i\), reaffirming that the quadratic formula not only provides solutions but helps verify the type of those solutions.