Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions of each equation. Do not use a calculator. $$x^{4}+x^{2}-6=0$$

Step-by-Step Solution

Answer

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

Step 1: Substitute a Variable

Let\( y = x^2 \). The given equation \( x^4 + x^2 - 6 = 0 \) then becomes \( y^2 + y - 6 = 0 \). This substitution simplifies the quartic equation into a quadratic one.

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \( y^2 + y - 6 = 0 \) by factoring or using the quadratic formula. Factor the equation as \( (y + 3)(y - 2) = 0 \).

Step 3: Find the Values of y

Setting each factor equal to zero gives the solutions for \( y \): \( y + 3 = 0 \) or \( y - 2 = 0 \). Thus, \( y = -3 \) and \( y = 2 \).

Step 4: Solve for x using y values

Since \( y = x^2 \), we substitute back to solve for \( x \):- For \( y = 2 \),\( x^2 = 2 \) gives solutions \( x = \sqrt{2} \) and \( x = -\sqrt{2} \).- For \( y = -3 \), since \( x^2 = -3 \), this gives solutions \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \).

Step 5: List all Complex Solutions

The complex solutions for the original equation, \( x^4 + x^2 - 6 = 0 \), are \( x = \sqrt{2}, x = -\sqrt{2}, x = i\sqrt{3}, x = -i\sqrt{3} \).

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. It always involves terms with powers of \( x \) up to a second degree.
The general nature of quadratic equations allows for solutions that can be real or complex numbers. To find these solutions, we usually use methods like factoring, completing the square, or the quadratic formula. Each method provides a different path to reach the same result.
In the problem we're considering, the starting equation \( x^4 + x^2 - 6 = 0 \) was converted to a quadratic form using substitution. This translates the equation into a simpler quadratic type, which can then be solved using standard techniques.

Substitution Method

The substitution method is a powerful technique to simplify equations, especially when dealing with higher-degree polynomials. In our original exercise, we substitute \( y = x^2 \). This re-defines the problem, making it easier to manage.

• Transformation: Original equation \( x^4 + x^2 - 6 = 0 \) becomes the quadratic equation \( y^2 + y - 6 = 0 \).
• Simplification: It makes the equation easier to solve by reducing the degree.
• Reversion: After solving the simpler equation, revert the substitution to find the original variable's values.

This approach is efficient and helps eliminate the potential complexities that arise from directly solving higher-degree polynomials.

Imaginary Numbers

Imaginary numbers emerge when we encounter negative numbers under a square root in equations. These numbers are expressed in terms of \( i \), where \( i = \sqrt{-1} \).
In this exercise, when we solved \( x^2 = -3 \), the solutions were complex because they involved imaginary numbers. The results were \( x = i\sqrt{3} \) and \( x = -i\sqrt{3} \).
Imaginary numbers extend the notion of numbers beyond the real line, allowing us to handle a wider range of equations. Understanding this concept is crucial for analyzing the roots of many complex polynomials.

Factoring Technique

Factoring is a method for solving polynomial equations by expressing them as a product of simpler polynomials whose solutions are easily determined. In the context of the substitution method, once the quadratic equation \( y^2 + y - 6 = 0 \) was identified, it was factored into \( (y + 3)(y - 2) = 0 \).

• Set each factor equal to zero: \( y + 3 = 0 \) gives \( y = -3 \) and \( y - 2 = 0 \) gives \( y = 2 \).
• Back-Substitute: Replacing \( y \) with \( x^2 \) as per the original substitution completes the solution.

Factoring is particularly effective for quadratic equations because it breaks them into linear factors, simplifying the solution process greatly.