Problem 69 - 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 \(-2 + 2\sqrt{2}i\) and \(-2 - 2\sqrt{2}i\).

Step 1: Rewrite Equation in Terms of z

The given equation \(z + 4 + \frac{12}{z} = 0\) can be rewritten by multiplying every term by \(z\) to eliminate the fraction. This gives us: \(z^2 + 4z + 12 = 0\).

Step 2: Use Quadratic Formula

The equation \(z^2 + 4z + 12 = 0\) is a quadratic equation. We can solve it using the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For our equation, \(a = 1\), \(b = 4\), and \(c = 12\).

Step 3: Calculate the Discriminant

Calculate the discriminant of the quadratic equation: \(b^2 - 4ac\). Substituting the values, we have \(4^2 - 4(1)(12) = 16 - 48 = -32\). The discriminant is negative, indicating complex solutions.

Step 4: Solve for z Using the Quadratic Formula

Substitute into the quadratic formula: \(z = \frac{-4 \pm \sqrt{-32}}{2}\). Simplifying further, we get \(z = \frac{-4 \pm \sqrt{32}i}{2}\).

Step 5: Simplify the Expression

Split and simplify: \(z = \frac{-4}{2} \pm \frac{\sqrt{32}i}{2}\). Simplifying \(\sqrt{32}\) gives \(4\sqrt{2}\), so \(z = -2 \pm 2\sqrt{2}i\). Therefore, the solutions are \(-2 + 2\sqrt{2}i\) and \(-2 - 2\sqrt{2}i\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and come in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). These equations are called quadratic because the highest power of the unknown, typically represented as \(x\) or \(z\) in equations, is two.
To solve a quadratic equation, the goal is usually to find the values of \(x\) that satisfy the equation. These values are known as the roots of the equation.

• If the equation can be factored easily, this method might be used.
• If not, methods like completing the square or the quadratic formula can be applied.

The equation in our problem started as a non-standard form, \(z + 4 + \frac{12}{z} = 0\), but was transformed into standard form by eliminating the fraction, yielding \(z^2 + 4z + 12 = 0\). This transformation was crucial because it allowed us to apply techniques specific to quadratic equations.

Discriminant

The discriminant is a key concept when solving quadratic equations. It is the part of the quadratic formula under the square root: \(b^2 - 4ac\).
The discriminant allows us to determine the nature of the equation's roots without solving the equation completely.

• If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root, also known as a repeated or double root.
• If \(b^2 - 4ac < 0\), the roots are complex or imaginary numbers.

In our example, the quadratic equation \(z^2 + 4z + 12 = 0\) has a discriminant \(16 - 48 = -32\). This negative result tells us that the solutions are complex numbers. These complex numbers appear in conjugate pairs, of the form \(a + bi\) and \(a - bi\). This insight is essential for predicting and understanding the type of solutions.

Quadratic Formula

One of the most powerful tools for solving quadratic equations is the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). It provides a straightforward way to find the roots of any quadratic equation, regardless of its complexity.
Applying the quadratic formula involves three main steps:

• Identify coefficients \(a\), \(b\), and \(c\) from the equation \(ax^2 + bx + c = 0\).
• Calculate the discriminant: \(b^2 - 4ac\).
• Plug these values into the formula and solve for \(x\).

In our scenario, the quadratic equation was \(z^2 + 4z + 12 = 0\), where \(a = 1\), \(b = 4\), and \(c = 12\). After computing the discriminant and confirming its negative value, inserting these into the quadratic formula gives us solutions with an imaginary component, specifically \(z = -2 \pm 2\sqrt{2}i\). This result is a clear demonstration of how the quadratic formula handles equations resulting in complex roots.