Problem 72 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$x^{2}+\frac{1}{2} x+1=0$$

Step-by-Step Solution

Answer

The solutions are \( x_1 = -\frac{1}{4} + i \cdot \frac{\sqrt{15}}{4} \) and \( x_2 = -\frac{1}{4} - i \cdot \frac{\sqrt{15}}{4} \).

Step 1: Identify the Type of Equation

We are given a quadratic equation of the form \( ax^2 + bx + c = 0 \), with coefficients \( a = 1 \), \( b = \frac{1}{2} \), and \( c = 1 \). The goal is to find the roots of this quadratic equation.

Step 2: Use the Quadratic Formula

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used to find the solutions for \( x \). Substituting in the values, we have: \[ x = \frac{-\left(\frac{1}{2}\right) \pm \sqrt{\left(\frac{1}{2}\right)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \] Simplifying gives: \[ x = \frac{-\frac{1}{2} \pm \sqrt{\frac{1}{4} - 4}}{2} \]

Step 3: Simplify the Discriminant

Calculate \( \frac{1}{4} - 4 \), which equals \( \frac{1}{4} - \frac{16}{4} = -\frac{15}{4} \). This is the discriminant, and since it is negative, we will have complex solutions.

Step 4: Simplify the Square Root of the Discriminant

The expression \( \sqrt{-\frac{15}{4}} \) simplifies to \( i \cdot \sqrt{\frac{15}{4}} = i \cdot \frac{\sqrt{15}}{2} \), due to the property of square roots of negative numbers involving \( i \), the imaginary unit.

Step 5: Substitute Back and Simplify

Substitute \( i \cdot \frac{\sqrt{15}}{2} \) back into the quadratic formula: \[ x = \frac{-\frac{1}{2} \pm i \cdot \frac{\sqrt{15}}{2}}{2} \] This simplifies further to: \[ x = -\frac{1}{4} \pm i \cdot \frac{\sqrt{15}}{4} \].

Step 6: Express Solutions in the Form \(a + bi\)

The solutions can be written in the standard form for complex numbers as: 1. \( x_1 = -\frac{1}{4} + i \cdot \frac{\sqrt{15}}{4} \) 2. \( x_2 = -\frac{1}{4} - i \cdot \frac{\sqrt{15}}{4} \) Both solutions are expressed as \( a + bi \).

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree 2, having the general form \( ax^2 + bx + c = 0 \). Here, "a," "b," and "c" are coefficients where "a" is not zero. The equation forms a parabola when graphed. Quadratic equations are central to algebra and can describe the motion of objects under uniform acceleration or find maximum and minimum values in various functions. Quadratic equations may have:

• Two real and distinct solutions.
• A single real solution (also known as a double root).
• Two complex solutions when no real solutions exist.

Understanding how to solve these equations is essential, as it allows for the exploration of different mathematical behaviors and real-world applications.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. Given any quadratic equation \( ax^2 + bx + c = 0 \), the formula provides the roots of the equation as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula allows us to find the solutions directly by inserting the coefficients "a," "b," and "c." It is especially useful when factoring is difficult or impossible. By using this formula:

• You can solve any quadratic equation, regardless of whether or not it can be factored easily.
• It highlights the role of the discriminant (\( b^2 - 4ac \)) in determining the nature of the roots.

The quadratic formula ensures that you can always find solutions, whether they are real or complex, forming an essential part of the algebra toolkit.

Discriminant

The discriminant of a quadratic equation is the part of the quadratic formula under the square root: \( b^2 - 4ac \). This value helps determine the nature of the roots of the quadratic equation. Here's how to interpret the discriminant:- If \( b^2 - 4ac > 0 \), there are two distinct real roots.- If \( b^2 - 4ac = 0 \), there is exactly one real root (also called a repeated or double root).- If \( b^2 - 4ac < 0 \), the roots are complex conjugates.In the given problem, the discriminant is negative (\( -\frac{15}{4} \)), leading to complex solutions. The discriminant offers insights into the relationship between the coefficients of the quadratic and its solutions, making it a key part of solving quadratic equations efficiently.

Complex Solutions

Complex solutions occur when the discriminant of a quadratic equation is negative, meaning it does not intersect the x-axis on a graph. Complex numbers take the form \( a + bi \), where "a" and "b" are real numbers, and "i" is the imaginary unit, defined by \( i = \sqrt{-1} \). In our example:

• The quadratic formula provides solutions by using the concept of imaginary numbers.
• The expression \( \sqrt{-1} \) is simplified using "i," allowing us to resolve the square root of a negative number.

The solutions \( x_1 = -\frac{1}{4} + i \cdot \frac{\sqrt{15}}{4} \) and \( x_2 = -\frac{1}{4} - i \cdot \frac{\sqrt{15}}{4} \) are both complex. By understanding these solutions, students gain insight into how quadratic equations can extend into the complex plane, opening up further mathematical exploration.