Problem 23 - A Graphical Approach to College Algebra

Question

Find all complex solutions for each equation by hand. Do not use a calculator. $$x^{-4}-3 x^{-2}-4=0$$

Step-by-Step Solution

Answer

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

Step 1: Substitution

Let us begin by substituting a new variable to make the equation simpler. Let \( y = x^{-2} \). That means \( y^2 = (x^{-2})^2 = x^{-4} \). This allows us to rewrite the original equation \( x^{-4} - 3x^{-2} - 4 = 0 \) as \( y^2 - 3y - 4 = 0 \).

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \( y^2 - 3y - 4 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -3, c = -4 \). Substitute these values into the formula: \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 1 \times (-4)}}{2 \times 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} \]. This gives two solutions: \( y = 4 \) and \( y = -1 \).

Step 3: Resubstitution to Original Variable

Replace \( y \) with \( x^{-2} \). The two cases are \( x^{-2} = 4 \) and \( x^{-2} = -1 \).

Step 4: Solve for x (Case 1, x^{-2} = 4)

If \( x^{-2} = 4 \), then take the reciprocal to get \( x^2 = \frac{1}{4} \). Solving this gives \( x = \pm \frac{1}{2} \). Therefore, the solutions for this case are \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

Step 5: Solve for x (Case 2, x^{-2} = -1)

If \( x^{-2} = -1 \), taking the reciprocal gives \( x^2 = -1 \). The solutions to this equation are complex numbers, \( x = i \) and \( x = -i \), where \( i \) is the imaginary unit.

Step 6: Conclusion

Combine all solutions found from both cases together. The complete set of complex solutions for the equation \( x^{-4} - 3x^{-2} - 4 = 0 \) is \( x = \frac{1}{2}, -\frac{1}{2}, i, -i \).

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a fundamental concept in mathematics. It is a polynomial equation of degree 2, generally taking the form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). Quadratic equations form a parabola when graphed. The solutions to a quadratic equation are the points where this parabola intersects the x-axis. These solutions can be real or complex.

• Real solutions: These occur when the discriminant \( b^2 - 4ac \) is positive or zero.
• Complex or imaginary solutions: These occur when the discriminant is negative. They include the imaginary unit \( i \), which has the property \( i^2 = -1 \).

To solve quadratic equations, various methods can be employed, such as factoring, completing the square, or using the quadratic formula. In our exercise, we used the quadratic formula, providing an efficient way to find solutions when factoring is difficult or impossible.

The Role of the Imaginary Unit

In mathematics, the imaginary unit is crucial for expanding the number system to include complex numbers. It is represented by \( i \) and defined by the property that \( i^2 = -1 \). This lets us solve equations involving negative values under the square root, such as \( x^2 = -1 \). When we solve such an equation, the solutions are \( x = i \) and \( x = -i \). These solutions form part of what we call complex solutions. Complex numbers consist of two parts: a real part and an imaginary part.

• Real part: For example, in \( 3 + 4i \), the real part is 3.
• Imaginary part: The imaginary part in \( 3 + 4i \) is 4i.

The imaginary unit makes it possible to model and solve a wide range of mathematical problems, especially those that do not resolve within the real number system. A complex solution like \( x = i \) emerged in our solution process where \( x^{2} = -1 \). Understanding this concept helps integrate the role of complex numbers in solving more complicated equations.

Exploring Variable Substitution

Variable substitution is a powerful technique often used to simplify complicated equations, especially those that are polynomial in nature. The essence of substitution involves replacing a variable or expression within an equation with another simpler variable or expression.In our original exercise, we dealt with the equation: \[ x^{-4} - 3x^{-2} - 4 = 0 \] To simplify this equation, we used the substitution \( y = x^{-2} \). This transformed the problem into a standard quadratic form: \[ y^2 - 3y - 4 = 0 \] This step made the equation easier to manage, allowing the application of the quadratic formula. After solving the quadratic form, re-substitution was necessary to revert to the original variable format.

• Initial substitution: Simplifies the equation by reducing complexity.
• Solving the equation: Using methods such as factoring, completing the square, or applying the quadratic formula.
• Re-substitution: Converts the solution back to the original problem's variable context.

Substitution not only simplifies complex algebraic expressions but also enhances understanding, paving pathways to more efficient problem-solving methods. It was critical in our ability to handle the original exercise effectively.