Problem 67 - Algebra and Trigonometry

Question

Find all solutions of the equation, and express them in the form \(a+b i\) $$ t+3+\frac{3}{t}=0 $$

Step-by-Step Solution

Answer

Solutions: \( t_1 = -\frac{3}{2} + \frac{\sqrt{3}}{2}i \) and \( t_2 = -\frac{3}{2} - \frac{\sqrt{3}}{2}i \).

Step 1: Simplify the Equation

Start by rewriting the given equation: \[ t + 3 + \frac{3}{t} = 0. \] Multiply the entire equation by \(t\) to eliminate the fraction:\[ t^2 + 3t + 3 = 0. \] This is a quadratic equation in the form \( at^2 + bt + c = 0 \) with \( a = 1, b = 3, \) and \( c = 3. \)

Step 2: Use the Quadratic Formula

The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used for finding the roots of the equation. Here, \( a = 1, b = 3, \) and \( c = 3. \)Insert these values into the formula:\[ t = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}. \] This simplifies to:\[ t = \frac{-3 \pm \sqrt{9 - 12}}{2}. \] \[ t = \frac{-3 \pm \sqrt{-3}}{2}. \]

Step 3: Simplify the Square Root of a Negative Number

We have \( \sqrt{-3} \), which is an imaginary number. By definition, \( \sqrt{-1} = i \). Thus, \( \sqrt{-3} = \sqrt{3} \times i \).Substitute this back into the equation:\[ t = \frac{-3 \pm \sqrt{3}i}{2}. \]

Step 4: Express in the Form \(a + bi\)

Simplify the expression \( t = \frac{-3 \pm \sqrt{3}i}{2} \) into the form \( t = a + bi \): - For the plus sign: \( t = -\frac{3}{2} + \frac{\sqrt{3}}{2}i \)- For the minus sign: \( t = -\frac{3}{2} - \frac{\sqrt{3}}{2}i \).Thus, the solutions are:\[ t_1 = -\frac{3}{2} + \frac{\sqrt{3}}{2}i \text{ and } t_2 = -\frac{3}{2} - \frac{\sqrt{3}}{2}i. \]

Key Concepts

Quadratic Equations

Quadratic equations form one of the fundamental topics in algebra. They are equations of the second degree, generally written as \( ax^2 + bx + c = 0 \). In our example, the equation \( t^2 + 3t + 3 = 0 \) is a quadratic equation. The degree of the equation is determined by the highest power of the variable or unknown, which is 2 in this case.
Quadratic equations can have two solutions, and these solutions can be real or complex numbers.
To solve quadratic equations, we often use methods such as:

• Factoring, when possible.
• Completing the square.
• Using the quadratic formula, which is applicable for all quadratic equations.

Understanding the structure and properties of quadratic equations makes it easier to determine the most suitable method for finding their solutions.

Imaginary Numbers

Imaginary numbers extend the concept of the number system. They are numbers that, when squared, yield a negative result. The most common imaginary unit is denoted as \( i \), where \( i^2 = -1 \).
This concept becomes important when dealing with the square root of negative numbers, as in our quadratic solution involving \( \sqrt{-3} \).
In our example, \( \sqrt{-3} \) is re-expressed using \( i \). Therefore, \( \sqrt{-3} = \sqrt{3}i \).
Imaginary numbers, when combined with real numbers, form complex numbers, typically expressed as \( a + bi \). Imaginary numbers are crucial in solving quadratic equations with negative discriminant when using the quadratic formula.

Quadratic Formula

The quadratic formula is a universal tool for solving any quadratic equation. It is expressed as:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides the roots of a quadratic equation \( ax^2 + bx + c = 0 \).
It works regardless of whether the solutions are real or complex. In our example, the values are \( a = 1 \), \( b = 3 \), and \( c = 3 \).
Substituting these values simplifies the quadratic formula to find the equation's roots. If the term under the square root, known as the discriminant \( b^2 - 4ac \), is negative, like in our case, it indicates the presence of imaginary solutions. Here, \( b^2 - 4ac = -3 \), leading to imaginary numbers and indicating that the equation has no real solutions.