Problem 68 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$z+4+\frac{12}{z}=0$$

Step-by-Step Solution

Answer

The solutions are \(-2 + 2i\sqrt{2}\) and \(-2 - 2i\sqrt{2}\).

Step 1: Simplify the Equation

The given equation is \(z + 4 + \frac{12}{z} = 0\). Subtract 4 from both sides to get \(z + \frac{12}{z} = -4\).

Step 2: Multiply Through by the Denominator

Multiply every term by \(z\) (assuming \(z eq 0\)) to eliminate the fraction: \(z^2 + 12 = -4z\).

Step 3: Rearrange to Form a Quadratic Equation

Reorder the equation to hold all terms in one side: \(z^2 + 4z + 12 = 0\).

Step 4: Use the Quadratic Formula

For the quadratic equation \(az^2 + bz + c = 0\), the solutions for \(z\) are given by the quadratic formula: \[z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here, \(a = 1\), \(b = 4\), and \(c = 12\). Substitute these values into the formula.

Step 5: Calculate the Discriminant

The discriminant \(b^2 - 4ac\) is \(4^2 - 4 \cdot 1 \cdot 12 = 16 - 48 = -32\). Since the discriminant is negative, the roots are complex.

Step 6: Compute the Roots

Using the quadratic formula: \[z = \frac{-4 \pm \sqrt{-32}}{2}\].First, simplify \(\sqrt{-32}\) to \(i\sqrt{32}\) or \(4i\sqrt{2}\). This gives:\[z = \frac{-4 \pm 4i\sqrt{2}}{2}\].Divide through by 2:\[z = -2 \pm 2i\sqrt{2}\].

Step 7: Express Solutions as a + bi

The solutions are \(-2 + 2i\sqrt{2}\) and \(-2 - 2i\sqrt{2}\). These are in the form \(a + bi\), with \(a = -2\) and \(b = \pm 2\sqrt{2}\).

Key Concepts

Quadratic Equation

The quadratic equation is one of the fundamental concepts in algebra. It takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In this exercise, after simplifying and rearranging, we derived the quadratic equation \(z^2 + 4z + 12 = 0\).
Quadratic equations can be solved through several methods including factoring, completing the square, and using the quadratic formula. However, not all quadratic equations are factorable, which makes the quadratic formula a universal method for finding solutions. The roots of a quadratic equation can be real or complex, depending on the value of the discriminant. Understanding the structure and solutions of quadratic equations is essential, as it lays the groundwork for solving more intricate polynomial equations.

Quadratic Formula

The quadratic formula provides a systematic way to find the roots of any quadratic equation. It is particularly helpful when factoring is challenging or impossible. The formula is \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(az^2 + bz + c = 0\).
Using the quadratic formula ensures that you can find solutions for any quadratic equation, with confidence. You simply plug the coefficients into the formula and solve. Even for complex coefficients, the quadratic formula remains consistent. As demonstrated in the original exercise, after substituting the values \(a = 1\), \(b = 4\), and \(c = 12\), we calculated the complex solutions, which included complex numbers as roots. This highlights the robustness of the quadratic formula in handling diverse types of equations.

Discriminant

The discriminant is a powerful component of the quadratic formula, indicated by \(b^2 - 4ac\). It reveals the nature of the roots without actually solving the equation. Depending on its value, the discriminant can tell us whether the roots are real or complex and whether they are distinct or repeated.

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, the roots are complex conjugates.

In the exercise above, we found the discriminant to be \(-32\), a negative value, which indicates complex solutions. Understanding the discriminant helps in anticipating the type of roots you can expect before fully solving the quadratic equation. Knowing this, you can prepare for dealing with complex numbers, as was required in this problem.