Problem 43 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (2 x-1)(8 x-4)=-1 $$

Step-by-Step Solution

Answer

x = \frac{1}{2} \pm \frac{i}{4}

Step 1: Expand the equation

First, expand the given equation \( (2x - 1)(8x - 4) = -1 \). Multiply the two binomials: \[ (2x - 1)(8x - 4) = 2x \times 8x + 2x \times (-4) + (-1) \times 8x + (-1) \times (-4) \] which simplifies to \[ 16x^2 - 8x - 8x + 4 = 16x^2 - 16x + 4. \] Therefore, the equation to solve becomes \[ 16x^2 - 16x + 4 = -1. \]

Step 2: Move all terms to one side

Move -1 to the left side of the equation to set it equal to 0: \[ 16x^2 - 16x + 4 + 1 = 0 \] which simplifies to \[ 16x^2 - 16x + 5 = 0. \]

Step 3: Identify coefficients for the quadratic formula

Identify the coefficients from the quadratic equation \( 16x^2 - 16x + 5 = 0 \). Here, \ a = 16 \, \ b = -16 \, and \ c = 5 \.

Step 4: Apply the quadratic formula

Use the quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Plug in the values of \ a \, \ b \, and \ c \: \[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 16 \times 5}}{2 \times 16} \] which simplifies to \[ x = \frac{16 \pm \sqrt{256 - 320}}{32}. \]

Step 5: Simplify the expression under the square root

Simplify the expression under the square root: \[ 256 - 320 = -64. \] Therefore, \[ x = \frac{16 \pm \sqrt{-64}}{32}. \]

Step 6: Simplify the square root of a negative number

Since \ \sqrt{-64} = 8i \, substitute back into the equation: \[ x = \frac{16 \pm 8i}{32} \] which simplifies to \[ x = \frac{16}{32} \pm \frac{8i}{32}, \] giving \[ x = \frac{1}{2} \pm \frac{i}{4}. \]

Key Concepts

Complex Numbers

Complex numbers are numbers that include both a real part and an imaginary part. The imaginary unit is represented by \( i \), where \( i^2 = -1 \). Complex numbers are written in the form \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part.
Here are a few key points to understand:

• When the quadratic formula yields a negative number under the square root, the solutions are complex numbers.
• For example, in our exercise, the term under the square root was \( 256 - 320 = -64 \). The square root of \( -64 \) became \( 8i \).
• This gave us solutions of the form \( x = \frac{1}{2} \text{±} \frac{i}{4} \). This means there are two solutions: \( \frac{1}{2} + \frac{i}{4} \) and \( \frac{1}{2} - \frac{i}{4} \).

Understanding complex numbers is crucial as they extend our number system and allow us to solve equations that have no real solutions.

Quadratic Equations

Quadratic equations are polynomial equations of degree 2 and can be written in the form \( ax^2 + bx + c = 0 \). The solutions to these equations can be found using various methods, with the quadratic formula being one of the most common and effective.
Important points about quadratic equations include:

• The standard form is \( ax^2 + bx + c = 0 \) where \( a, b, \) and \( c \) are constants.
• The quadratic formula, \( x = \frac{-b \text{±} \frac{b^2 - 4ac}}{2a} \), can solve any quadratic equation.
• In our exercise, after expanding and rearranging the equation \( (2x - 1)(8x - 4) = -1 \), we obtained \( 16x^2 - 16x + 5 = 0 \).

Quadratic equations appear in various contexts in algebra and are foundational for understanding more advanced topics in mathematics.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. These expressions can be simplified, factored, and manipulated using algebraic rules.
Key points about algebraic expressions include:

• They can be operations like addition, subtraction, multiplication, and division.
• In our exercise, we started with the algebraic expression \( (2x - 1)(8x - 4) \). Multiplying these binomials gave us \( 16x^2 - 16x + 5 \).
• Simplifying algebraic expressions is often the first step in solving complex equations.

Mastering the manipulation of algebraic expressions is vital for solving equations, including quadratic and higher-order equations.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. For quadratic equations, the quadratic formula is a powerful tool.
Steps for solving the given quadratic equation included:

• First, we expanded \( (2x - 1)(8x - 4) \) and simplified it to \( 16x^2 - 16x + 5 = 0 \).
• Next, we identified coefficients: \( a = 16 \), \( b = -16 \), and \( c = 5 \).
• Using the quadratic formula \( x = \frac{-b \text{±} \frac{b^2 - 4ac}}{2a} \), we solved for \( x \).

This process demonstrates how systematic steps and the quadratic formula can be applied to find solutions, even complex ones.