Problem 62 - Algebra and Trigonometry

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: Move Terms to One Side

Starting with the equation \( z + 4 + \frac{12}{z} = 0 \), subtract 4 from both sides to isolate the terms with \( z \) on one side:\[z + \frac{12}{z} = -4\]

Step 2: Eliminate the Fraction

Multiply every term by \( z \) to eliminate the fraction:\[z^2 + 12 = -4z\] Reorganize the equation to form a standard quadratic equation:\[z^2 + 4z + 12 = 0\]

Step 3: Identify Coefficients

Identify the coefficients of the quadratic equation \( z^2 + 4z + 12 = 0 \).The standard form is \( az^2 + bz + c = 0 \).- \( a = 1 \)- \( b = 4 \)- \( c = 12 \)

Step 4: Calculate the Discriminant

The discriminant \( \Delta \) is calculated using the formula \( \Delta = b^2 - 4ac \):\[\Delta = 4^2 - 4 \times 1 \times 12 = 16 - 48 = -32\]A negative discriminant indicates the solutions are complex.

Step 5: Apply the Quadratic Formula

Use the quadratic formula to find the roots:\[ z = \frac{-b \pm \sqrt{\Delta}}{2a} \]Substitute \( a = 1 \), \( b = 4 \), and \( \Delta = -32 \) into the formula:\[ z = \frac{-4 \pm \sqrt{-32}}{2} \]

Step 6: Simplify the Square Root of a Negative Number

Express \( \sqrt{-32} \) as \( \sqrt{32} i \). Recall \( \sqrt{-1} = i \):\[\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \]Thus, \( \sqrt{-32} = 4\sqrt{2} i \).

Step 7: Solve for Complex Solutions

Substitute \( \sqrt{-32} = 4\sqrt{2} i \) into the quadratic formula:\[z = \frac{-4 \pm 4\sqrt{2} i}{2} = \frac{-4}{2} \pm \frac{4\sqrt{2} i}{2}\]Simplify:\[z = -2 \pm 2\sqrt{2} i\]These are the complex solutions expressed in the form \( a + bi \).

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). These equations are called 'quadratic' because they involve terms up to the second degree, which means the maximum exponent on the variable is 2. Quadratic equations can appear in many situations, including physics, engineering, and even finance.

• They can have two solutions because of their parabolic graph.
• The solutions to these equations are where the graph intersects the x-axis.

Sometimes they come in forms that are not immediately recognizable as quadratics, and terms might need rearranging or simplifying, like in the original exercise, where elimination of fractions turned the equation into a standard quadratic form.

Discriminant

The discriminant is a key component in understanding the nature of solutions for quadratic equations. It is represented by \( \Delta \) and is calculated using the formula \( \Delta = b^2 - 4ac \). The value of the discriminant tells us:

• If \( \Delta > 0 \), the quadratic equation has two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (or a repeated root).
• If \( \Delta < 0 \), the equation has two complex conjugate solutions.

In the exercise given, the discriminant was calculated to be \( -32 \), which is less than zero. This immediately tells us that the solutions will be complex, not real. The discriminant provides insight even before finding the exact solutions.

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions of any quadratic equation. The formula is written as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows us to find the roots of the quadratic equation using substitutions of the coefficients \( a \), \( b \), and \( c \) along with the discriminant. In the exercise, the coefficients used were \( a = 1 \), \( b = 4 \), and \( c = 12 \).

By applying these in the formula, you can derive the solutions. Taking the square root of a negative number under the formula led to complex numbers, underlining the flexibility and completeness of this method in handling any quadratic function.

Complex Solutions

Complex solutions occur when the discriminant of a quadratic equation is negative, as was the case in the given exercise. Complex numbers involve the imaginary unit \( i \), where \( i^2 = -1 \), which allows for the representation of numbers like \( \sqrt{-32} \) as \( 4\sqrt{2}i \).

Complex solutions generally come in pairs of complex conjugates, such as \( -2 + 2\sqrt{2}i \) and \( -2 - 2\sqrt{2}i \). These numbers are the outputs for the quadratic formula where \( \Delta < 0 \), demonstrating that every quadratic equation has two solutions, even if they are not real. This concept of complex solutions broadens the scope of mathematics, allowing it to handle a wider range of problems.