Problem 49 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate. $$3 n^{2}-6 n+4=0$$

Step-by-Step Solution

Answer

The solutions are \( n = 1 + \frac{i\sqrt{3}}{3} \) and \( n = 1 - \frac{i\sqrt{3}}{3} \).

Step 1: Determine the Method

The equation of the form \( an^2 + bn + c = 0 \) is a quadratic equation. We will use the quadratic formula to solve this, as it is not factorable at first glance. The quadratic formula is \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 2: Identify Coefficients

In the quadratic equation \( 3n^2 - 6n + 4 = 0 \), the coefficients are: \( a = 3 \), \( b = -6 \), \( c = 4 \).

Step 3: Calculate the Discriminant

The discriminant \( D \) is calculated using \( D = b^2 - 4ac \). Substituting the values gives \( D = (-6)^2 - 4 \times 3 \times 4 = 36 - 48 = -12 \).

Step 4: Interpret the Discriminant

The discriminant is \(-12\), which is less than zero, indicating that the quadratic equation has no real solutions and two complex solutions.

Step 5: Calculate the Complex Solutions

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( b^2 - 4ac = -12 \), we have:\[n = \frac{-(-6) \pm \sqrt{-12}}{2 \times 3} = \frac{6 \pm \sqrt{12i^2}}{6} = \frac{6 \pm 2i\sqrt{3}}{6} = 1 \pm \frac{i\sqrt{3}}{3}.\]Thus, the solutions are \( n = 1 + \frac{i\sqrt{3}}{3} \) and \( n = 1 - \frac{i\sqrt{3}}{3} \).

Key Concepts

Quadratic Formula

The quadratic formula is one of the key tools in algebra for finding the roots of quadratic equations. A quadratic equation typically looks like this: \( ax^2 + bx + c = 0 \). When solving such equations, the quadratic formula provides an efficient method to find the solutions, or roots, directly. The formula is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \), \( b \), and \( c \) are the coefficients of the equation. It's crucial because it works even when the equation can’t be easily factored.
This broad applicability makes it a fundamental tool to know in algebra. The term under the square root sign, \( b^2 - 4ac \), is called the discriminant, which plays a vital role in determining the nature of the solutions. Let's explore further how it is used to find both real and complex roots.

Complex Solutions

Quadratic equations can have different types of solutions, which depend on the value of the discriminant. If the discriminant is negative, it indicates the solutions are not real numbers—they are complex numbers.
Complex solutions are written in the form of \( a + bi \), where \( i \) is the imaginary unit, satisfying \( i^2 = -1 \). This means the solutions have both a real and an imaginary part.

• In the exercise, the quadratic equation has a discriminant of -12, showing it has two complex solutions.
• These solutions are computed using the quadratic formula, substituting \( D \) with the negative value.
• The complex roots in this scenario are \( n = 1 + \frac{i\sqrt{3}}{3} \) and \( n = 1 - \frac{i\sqrt{3}}{3} \).

Understanding complex solutions requires comfort with operations involving imaginary numbers, which are essential for applications in engineering and physics.

Discriminant

The discriminant of a quadratic equation is derived from the formula \( D = b^2 - 4ac \). It is a powerful number because it tells us about the nature of the roots of the quadratic equation.
Here's what the discriminant can reveal:

• If \( D > 0 \), there are two distinct real solutions.
• If \( D = 0 \), there is exactly one real solution, also called a repeated or double root.
• If \( D < 0 \), the solutions are complex, meaning they have imaginary parts.

In this exercise, the discriminant was calculated to be \(-12\), which is less than zero. Thus, it confirms that the solutions are not real numbers but complex. This insight lets us pivot from expecting numeric solutions towards using complex number operations.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra that unlocks the ability to handle various mathematical problems. There are multiple methods to solve quadratic equations, including factoring, using the quadratic formula, completing the square, and graphing.
Here, we emphasize using the quadratic formula, especially when equations are not easily factorable.

• First, identify \( a \), \( b \), and \( c \) from the quadratic equation.
• Compute the discriminant \( b^2 - 4ac \) to determine the nature of the solutions.
• Apply the quadratic formula to find the roots, whether they are real or complex.

This systematic approach not only helps in handling textbook problems but also in real-world applications where quadratic relationships exist. Mastering this topic allows students to delve deeper into advanced topics, laying a solid groundwork in mathematical analysis.