Problem 34 - 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. $$\begin{aligned}&36 x^{4}+85 x^{2}+9=0\\\&[-4,4] \text { by }[-10,40]\end{aligned}$$

Step-by-Step Solution

Answer

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

Step 1: Set up the substitution

To simplify the quartic equation, we'll use a substitution. Let \( u = x^2 \). Then \( x^4 = u^2 \), so the equation becomes \( 36u^2 + 85u + 9 = 0 \). This is now a quadratic equation in terms of \( u \).

Step 2: Use the quadratic formula

The quadratic equation \( 36u^2 + 85u + 9 = 0 \) can be solved using the quadratic formula: \( u = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \), where \( a = 36 \), \( b = 85 \), and \( c = 9 \). Substituting, we get: \[ u = \frac{{-85 \pm \sqrt{{85^2 - 4 \cdot 36 \cdot 9}}}}{72} \].

Step 3: Calculate the discriminant and simplify

Calculate \( 85^2 - 4 \cdot 36 \cdot 9 \) to find the discriminant. This computes to \( 7225 - 1296 = 5929 \). The square root of \( 5929 \) is \( 77 \). Using these results, the quadratic formula yields \[ u = \frac{{-85 \pm 77}}{72} \].

Step 4: Solve for u

From the quadratic formula: \( u_1 = \frac{{-85 + 77}}{72} = \frac{{-8}}{72} = -\frac{1}{9} \) and \( u_2 = \frac{{-85 - 77}}{72} = \frac{{-162}}{72} = -\frac{9}{4} \). These are the values for \( u \) (i.e., \( x^2 \)).

Step 5: Solve for x

Since \( x^2 = u \), we have two equations: \( x^2 = -\frac{1}{9} \) and \( x^2 = -\frac{9}{4} \). Solving these gives the complex solutions: \( x = \pm i\frac{1}{3} \) and \( x = \pm i\frac{3}{2} \), where \( i \) is the imaginary unit.

Key Concepts

Quartic Equation

A quartic equation is a polynomial equation of degree four, meaning it is the highest degree polynomial that can still be solved by radicals, due to its end power of four. These types of equations can appear complicated because of their high degree, but they often can be simplified by substituting variables.

In this exercise, the quartic equation is transformed into a more manageable form by using the substitution method. By setting \( u = x^2 \), the equation \( 36x^4 + 85x^2 + 9 = 0 \) is simplified to the quadratic equation \( 36u^2 + 85u + 9 = 0 \).

• This transformation reduces the complexity, allowing us to apply techniques specific to quadratic equations.
• Once the solutions for \( u \) are obtained, they can be reverted back to solutions for \( x \).

It's a strategic way to break down problems that might otherwise look too intimidating to solve by directly calculating.

Quadratic Formula

The quadratic formula is a tried and true method for finding the roots of a quadratic equation. A quadratic equation is typically expressed in the form \( ax^2 + bx + c = 0 \). The quadratic formula itself is \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \).

Using this formula is extremely helpful because it provides a systematic approach to solving for the unknowns. In the context of our exercise, it was applied to solve \( 36u^2 + 85u + 9 = 0 \).

• Here, \( a = 36 \), \( b = 85 \), and \( c = 9 \).
• It involves substituting these values into the formula, which helps in calculating both the discriminant and the roots directly.

It's a powerful tool because it guarantees finding the solutions provided they exist, and it can also indicate when no real solutions are present.

Complex Numbers

Complex numbers extend the idea of numbers to include the square root of negative numbers. They are expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit, defined by \( i^2 = -1 \).

In this problem, the solutions for \( x \) are complex because the values of \( u \) were negative, leading to negative results when setting up the equation \( x^2 = u \).

• For example, \( x^2 = -\frac{1}{9} \) results in the solutions \( x = \pm i\frac{1}{3} \).
• Similarly, \( x^2 = -\frac{9}{4} \) translates to \( x = \pm i\frac{3}{2} \).

This concept is crucial when dealing with quadratic roots that are negative as it allows us to work beyond the limits of real numbers and access a broader field of solutions.

Discriminant

The discriminant is an important part of solving quadratic equations using the quadratic formula. Given a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is expressed as \( b^2 - 4ac \).

The value of the discriminant provides essential information:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (a repeated root).
• If \( \Delta < 0 \), there are two complex solutions, which was the case in our scenario as the solutions involved imaginary numbers.

In this exercise, the discriminant calculated was \( 5929 \), and since it was a perfect square, \( 77 \), it allowed clean computations leading to complex roots when it was substituted into the quadratic formula. Understanding this concept is key to predicting the nature of the solutions without necessarily solving the equation completely.