Problem 67 - Precalculus Mathematics for Calculus

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

The solutions are \( t = -\frac{3}{2} + \frac{i \sqrt{3}}{2} \) and \( t = -\frac{3}{2} - \frac{i \sqrt{3}}{2} \).

Step 1: Simplify and Rearrange the Equation

The original equation is \( t + 3 + \frac{3}{t} = 0 \).We start by eliminating the fraction by multiplying the entire equation by \( t \) to get:\[ t^2 + 3t + 3 = 0 \].

Step 2: Identify and Use the Quadratic Formula

The equation \( t^2 + 3t + 3 = 0 \) is quadratic with coefficients \( a = 1 \), \( b = 3 \), and \( c = 3 \). To find the solution, we use the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

Step 3: Calculate the Discriminant

First, calculate the discriminant \( \Delta = b^2 - 4ac \):\[ \Delta = 3^2 - 4 \times 1 \times 3 = 9 - 12 = -3 \].Since \( \Delta < 0 \), the solutions will be complex numbers.

Step 4: Solve for the Roots Using the Quadratic Formula

Using the quadratic formula, substitute the values for \( a \), \( b \), and the discriminant:\[ t = \frac{-3 \pm \sqrt{-3}}{2} \].This becomes: \[ t = \frac{-3}{2} \pm \frac{\sqrt{-3}}{2} \].

Step 5: Simplify the Complex Numbers

Simplifying further, we recognize \( \sqrt{-3} = i \sqrt{3} \). So the solutions become:\[ t = -\frac{3}{2} \pm \frac{i \sqrt{3}}{2} \].The solutions are in the form \( a + bi \).

Key Concepts

Quadratic Equations

Quadratic equations have a standard form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) represents the variable. These equations are called quadratic because the highest power of the variable is 2.
For example, in the equation \( t^2 + 3t + 3 = 0 \), the term \( t^2 \) makes it a quadratic equation.

• The term \( a \) is the coefficient of \( t^2 \), affecting the curvature.
• \( b \) is the coefficient of \( t \), affecting the slope.
• \( c \) is the constant term, affecting the vertical shift.

Quadratic equations can often be solved by factoring, completing the square, or using the quadratic formula.
The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a universal method that works for all quadratic equations, even when factoring isn't possible. In the given solution, we use the quadratic formula because the equation does not factor easily.

Complex Solutions

Complex solutions occur when the discriminant (\( \Delta = b^2 - 4ac \)) of a quadratic equation is negative. In the exercise equation \( t^2 + 3t + 3 = 0 \), the discriminant \( \Delta = -3 \) is negative. This leads us to complex solutions since there is no real square root for negative numbers.
Complex numbers are expressed in the form \( a + bi \), where:

• \( a \) is the real part.
• \( bi \) is the imaginary part, with \( i \) being the imaginary unit \( i = \sqrt{-1} \).

The quadratic formula handles complex solutions by representing the square root of the negative discriminant as an imaginary number. Therefore, the solution becomes \( t = \frac{-3}{2} \pm \frac{i \sqrt{3}}{2} \), representing two complex conjugate solutions: \( -\frac{3}{2} + \frac{i \sqrt{3}}{2} \) and \( -\frac{3}{2} - \frac{i \sqrt{3}}{2} \).

Imaginary Numbers

Imaginary numbers introduce a way to deal with the square roots of negative numbers, and the imaginary unit \( i \) is defined as \( i = \sqrt{-1} \). This allows us to rewrite the square roots of negative numbers in terms of \( i \).
For example, in the given step-by-step solution, \( \sqrt{-3} \) is rewritten as \( i \sqrt{3} \). This results in the final solutions being complex numbers that include imaginary terms.

• An imaginary term exists when the square root of a negative number is part of a mathematical expression.
• Imaginary numbers extend the real number system to the complex number system, which is crucial in solving many algebraic problems that don't have real solutions.

Using imaginary numbers, the concept of complex numbers emerges, which combines real and imaginary parts to form a comprehensive number system.
This is key in solutions like our exercise, where traditional real-number solutions don't apply, ensuring completeness in mathematical descriptions.