Problem 26 - Elementary Algebra

Question

Solve the following quadratic equations. \((8 d-6)^{2}=-24\)

Step-by-Step Solution

Answer

d = \frac{3 \pm i \sqrt{6}}{4}

Step 1: Setup the equation

Given the quadratic equation: \[(8d - 6)^2 = -24\]

Step 2: Take the square root of both sides

To isolate \(8d - 6\), take the square root of both sides of the equation: \[\sqrt{(8d - 6)^2} = \sqrt{-24}\]This simplifies to: \[8d - 6 = \pm\sqrt{-24}\]

Step 3: Simplify the square root

Since \(\sqrt{-24}\) includes an imaginary number, we can rewrite it as:\[\sqrt{-24} = \sqrt{-1 \cdot 24} = \sqrt{-1} \cdot \sqrt{24} = i \sqrt{24}\]Thus, we have: \[8d - 6 = \pm i \sqrt{24}\]

Step 4: Solve for d

Isolating \(d\) involves solving the equation: \[8d = 6 \pm i \sqrt{24}\]Finally, divide by 8:\[d = \frac{6 \pm i \sqrt{24}}{8}\]Simplify \(\sqrt{24}\) to \(2\sqrt{6}\) and then simplify the division:\[d = \frac{6 \pm 2i \sqrt{6}}{8} = \frac{3 \pm i \sqrt{6}}{4}\]

Key Concepts

solving quadratic equations

Solving quadratic equations involves finding the values of the variable that make the equation true. A standard quadratic equation is typically written in the form \[ax^2 + bx + c = 0\].
To solve such equations, you can use the quadratic formula, factoring, or completing the square. In our exercise, we started with: \[(8d - 6)^2 = -24\].
This equation is already in a form that makes squaring straightforward.

imaginary numbers

Imaginary numbers are numbers that give a negative result when squared. The basic unit of imaginary numbers is \(i\), which is defined as \(\sqrt{-1}\).
In the equation: \(\sqrt{(8d - 6)^2} = \sqrt{-24}\), solving \(\sqrt{-24}\) involves recognizing the presence of imaginary numbers. This becomes: \(\sqrt{-24} = i \sqrt{24}\), where \(i\) indicates an imaginary number. Remember, \(i^2 = -1\), which is fundamental to working with complex calculations involving imaginary numbers.

simplifying square roots

Simplifying square roots is a crucial step in solving quadratic equations, especially when dealing with complex numbers. Let's revisit: \(\sqrt{24}\). Square roots can be simplified by factoring the number under the square root into its prime factors.
For instance, \(24\) can be factored into \(4 \times 6\), so \(\sqrt{24}\) becomes \(\sqrt{4 \times 6} = \sqrt{4} \sqrt{6} = 2 \sqrt{6}\).
Breaking down \(\sqrt{24}\) makes manipulation easier and helps in further steps where the simplified forms play a crucial role in obtaining the final answer.

complex solutions

Complex solutions to quadratic equations occur when the solutions include imaginary numbers.\(Complex\) numbers are of the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
For our equation, we ended up with:\(d = \frac{6 \pm i\sqrt{24}}{8}\), which simplified to\(d = \frac{3 \pm i \sqrt{6}}{4}\).
Here,\(3/4\) is the real part, and\(i\frac{\sqrt{6}}{4}\) is the imaginary part. Understanding how to manage and simplify these terms is essential for mastering quadratic equations that yield complex solutions.