Problem 22 - 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. $$\begin{aligned}&9 x^{4}+35 x^{2}-4=0\\\&[-3,3] \text { by }[-10,100]\end{aligned}$$

Step-by-Step Solution

Answer

The complex solutions are \(x = \{ \frac{1}{3}, -\frac{1}{3}, 2i, -2i \} \).

Step 1: Identify a substitution

Recognize that the equation is quadratic in form for the term \(x^2\). Set \(u = x^2\), then rewrite the equation as \(9u^2 + 35u - 4 = 0\).

Step 2: Solve the quadratic equation

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(u\). Here, \(a = 9\), \(b = 35\), and \(c = -4\). Substitute these values to find \( u = \frac{-35 \pm \sqrt{35^2 - 4 \cdot 9 \cdot (-4)}}{2 \cdot 9}\).

Step 3: Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac\). For the given quadratic, it is \(1225 + 144 = 1369\). The square root of this discriminant is \(\sqrt{1369} = 37\).

Step 4: Solve for 'u'

Substitute the square root of the discriminant back into the quadratic formula:\[ u = \frac{-35 \pm 37}{18} \].Calculate both solutions: 1. \(u_1 = \frac{-35 + 37}{18} = \frac{2}{18} = \frac{1}{9} \)2. \(u_2 = \frac{-35 - 37}{18} = \frac{-72}{18} = -4 \). Thus, the possible values for \(u\) are \(\frac{1}{9}\) and \(-4\).

Step 5: Substitute back to find 'x'

Recall that \(u = x^2\). Solve for \(x\) for each value of \(u\):1. If \(u = \frac{1}{9}\), then \(x^2 = \frac{1}{9}\), so \(x = \pm \frac{1}{3}\).2. If \(u = -4\), there are no real solutions for \(x\), but in the complex plane, \(x = \pm i\sqrt{4} = \pm 2i\).

Step 6: Solution set

Combine the results to form the solution set:\(x = \{ \frac{1}{3}, -\frac{1}{3}, 2i, -2i \} \).

Step 7: Graph and Support

Graph the function \(Y_1 = 9x^4 + 35x^2 - 4\) using a graphing calculator with the viewing window [-3,3] by [-10,100]. Verify that \(x = \frac{1}{3}\) and \(x = -\frac{1}{3}\) are real roots by observing the points where the graph intersects the x-axis in this range.

Key Concepts

Quadratic Equation

The quadratic equation is a fundamental tool in algebra that takes the form \(ax^2 + bx + c = 0\). In this equation, \(a\), \(b\), and \(c\) are constants, with \(a \eq 0\). Quadratic equations are vital because they can be used to model various real-world situations, from physics to finance.
In the case of the original exercise, the equation was not immediately apparent as quadratic. By recognizing the substitution \(u = x^2\), we transformed the given polynomial into a quadratic form: \(9u^2 + 35u - 4 = 0\). This transformation is crucial as it simplifies solving highly complex polynomial equations by reducing them to quadratic ones.

Discriminant

The discriminant is a key component of the quadratic formula, given by \(b^2 - 4ac\). It provides critical information about the nature of the roots of a quadratic equation.
Determining the value of the discriminant allows us to know:

• If the roots are real or complex
• If the roots are distinct or repeated

For the exercise, the discriminant calculated as 1369 was a perfect square (37), indicating that there are two distinct real roots. This information is pivotal in understanding that different values for \(u\), such as obtained by the calculations, can exist in different number spaces, real or complex.

Graphing Calculator

Using a graphing calculator is advantageous for visualizing complex mathematical functions and confirming analytical solutions, especially for quadratic equations.
With the graphing calculator, the original function \(Y_1 = 9x^4 + 35x^2 - 4\) was graphed in the specified window of \([-3,3]\) by \([-10,100]\). The graph helps verify real solutions by intersecting points on the x-axis. Thus, \(x = \rac{1}{3}\) and \(x = -rac{1}{3}\) were observed as the real solutions.
Using the graph enhances understanding by providing a visual context for both the real and imaginary roots found through analytical solutions.

Real and Imaginary Numbers

In mathematics, numbers can be classified as either real or imaginary. Real numbers include any number on the number line but imaginary numbers arise when solutions involve the square root of negative numbers.
The exercise encountered complex solutions for \(x\), resulting in real numbers (\(rac{1}{3}\) and \(-rac{1}{3}\)) and imaginary numbers (\(2i\) and \(-2i\)).

• Real roots are solutions that can be seen on the graph as intersections.
• Imaginary roots, derived from \(x = \pm i \sqrt{4}\), correspond to non-intersecting points on the real part of the axis but are crucial in the complete solution set.

Understanding both kinds of numbers expands your grasp of complex operations and enriches overall mathematical competence.