Problem 35 - Intermediate Algebra

Question

Use the method of completing the square to solve each quadratic equation. $$ 3 n^{2}-6 n+5=0 $$

Step-by-Step Solution

Answer

No real solutions; solutions are complex.

Step 1: Move Constant Term

We start by moving the constant term to the other side of the equation to isolate the quadratic and linear terms. This gives us:\[ 3n^2 - 6n = -5 \]

Step 2: Factor Out the Leading Coefficient

Next, factor out the leading coefficient from the quadratic and linear terms:\[ 3(n^2 - 2n) = -5 \]

Step 3: Complete the Square

To complete the square, take half of the coefficient of the linear term inside the parentheses, square it, and add it inside the square:\[ n^2 - 2n \rightarrow (n - 1)^2 - 1 \]Then:\[ 3((n - 1)^2 - 1) = -5 \]

Step 4: Expand and Simplify

Distribute and simplify:\[ 3(n - 1)^2 - 3 = -5 \]Add 3 to both sides:\[ 3(n - 1)^2 = -2 \]

Step 5: Solve for the Square

Divide by 3 to isolate the square:\[ (n - 1)^2 = -\frac{2}{3} \]

Step 6: Determine Solutions

Solving \( (n - 1)^2 = -\frac{2}{3} \) presents no real number solutions because the expression inside the square root becomes negative, indicating complex solutions.

Key Concepts

Quadratic Equation

Quadratic equations are fundamental in algebra and appear in the form \( ax^2 + bx + c = 0 \). These equations can be solved using a variety of methods, such as factoring, using the quadratic formula, or completing the square, as in our exercise. Completing the square is a technique that involves transforming the quadratic equation into a perfect square trinomial.

In our example, the quadratic equation is \( 3n^2 - 6n + 5 = 0 \). The first step is to move the constant term \( 5 \) to the other side, resulting in \( 3n^2 - 6n = -5 \). This rearrangement is crucial to focus on the quadratic and linear terms only.

Next, the leading coefficient \( 3 \) is factored out from the quadratic and linear terms, simplifying the equation further to \( 3(n^2 - 2n) = -5 \). With these preparations, the equation is ready for completing the square, a vital vertex form conversion for solving.

Complex Solutions

Complex solutions emerge when solving a quadratic equation results in taking the square root of a negative number. In our exercise, the scenario arises as we reach the step \( (n - 1)^2 = -\frac{2}{3} \), where the negative sign indicates no real numbers, only complex solutions.

Complex numbers include a real part and an imaginary part, denoted as \( i \), where \( i^2 = -1 \). Thus, the expression \( \sqrt{-1} \) becomes \( i \).

To find complex solutions, take the square root of both sides of the equation. For \( (n - 1)^2 = -\frac{2}{3} \), the square root yields two solutions: \( n - 1 = \pm i\sqrt{\frac{2}{3}} \). Adding 1 to both solutions, the final solutions become \( n = 1 \pm i\sqrt{\frac{2}{3}} \). This showcases the necessity of understanding imaginary components in algebra.

Intermediate Algebra

Intermediate algebra bridges the gap between basic arithmetic and more advanced mathematical concepts. It prepares students for calculus and other higher math courses by introducing a variety of algebraic techniques, such as solving quadratic equations.

In the realm of intermediate algebra, understanding different methods to solve quadratics, like completing the square, is crucial. This method requires manipulation to turn the equation into a format that is easier to solve. Taking the example from our exercise, notice how reshaping the equation \( 3(n^2 - 2n) = -5 \) into a perfect square \( (n - 1)^2 \) involves intermediate algebra skills.

Grasping these techniques deeply enhances a student's ability to tackle complex algebraic problems. Intermediate algebra doesn't only involve problem-solving but also develops logical thinking, making it an essential part of a student's mathematical education.