Problem 66 - Algebra and Trigonometry

Question

Find all solutions of the equation, and express them in the form \(a+b i\) $$ 2 x^{2}+3=2 x $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}i \).

Step 1: Move all terms to one side of the equation

First, start with the equation: \[ 2x^2 + 3 = 2x \]Subtract \(2x\) from both sides to bring all terms to one side:\[ 2x^2 - 2x + 3 = 0 \]

Step 2: Identify the quadratic formula components

To solve the quadratic equation \( ax^2 + bx + c = 0 \), the coefficients are needed:\[ a = 2, \quad b = -2, \quad c = 3 \]

Step 3: Apply the quadratic formula

The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the coefficients:\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times 3}}{2 \times 2} \]

Step 4: Simplify the discriminant

Compute the value under the square root, known as the discriminant:\[ (-2)^2 - 4 \times 2 \times 3 = 4 - 24 = -20 \]Since the discriminant is negative, the solutions will be complex.

Step 5: Calculate complex solutions

Since the discriminant is negative, use \( \sqrt{-20} = \sqrt{20}i \):\[ x = \frac{2 \pm \sqrt{20}i}{4} \]Simplify \( \sqrt{20} \):\[ \sqrt{20} = 2\sqrt{5} \]Thus, the solutions are:\[ x = \frac{2 \pm 2\sqrt{5}i}{4} = \frac{1 \pm \sqrt{5}i}{2} \]

Step 6: Express the solutions in the form \(a + bi\)

The solutions are:\[ x = \frac{1}{2} + \frac{\sqrt{5}}{2}i \text{ and } x = \frac{1}{2} - \frac{\sqrt{5}}{2}i \]Both solutions are now in the standard form \( a + bi \).

Key Concepts

Quadratic Formula

When faced with a quadratic equation, which is an equation of the form \( ax^2 + bx + c = 0 \), the quadratic formula is your reliable tool for finding solutions. The quadratic formula is:

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

This formula provides solutions for any quadratic equation by using the coefficients \(a\), \(b\), and \(c\) from the equation. The "\( \pm \)" sign indicates that there are generally two solutions to a quadratic equation, since one involves addition and the other involves subtraction.
The beauty of the quadratic formula is that it works in all cases—whether the solutions are real or complex.

Discriminant

A crucial part of the quadratic formula solution process is calculating the discriminant, \( b^2 - 4ac \). The discriminant tells you about the nature of the roots of the quadratic equation:

• If the discriminant is positive, you have two distinct real solutions.
• If it equals zero, there is exactly one real solution (a double root).
• If the discriminant is negative, like in our case of \(-20\), this indicates that the solutions are not real numbers but complex numbers.

The discriminant essentially reveals whether the parabola described by the quadratic equation crosses the x-axis, touches it, or does not touch it at all. When the discriminant is negative, the solutions involve the square root of a negative number, leading directly to complex solutions.

Complex Solutions

Complex solutions arise when the quadratic equation has a negative discriminant. Complex numbers are written in the form \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part. Here, \( i \) is the imaginary unit, with the property \( i^2 = -1 \).
Given a negative discriminant, we compute the square root of a negative number in terms of \( i \). For example, if \( \sqrt{-20} \) appears in your calculations, you can rewrite it as \( \sqrt{20}i \). This involves calculating \( \sqrt{20} \) and then appending \( i \) to express it correctly in the complex form.

• This is why the solutions to the problem are \( \frac{1}{2} + \frac{\sqrt{5}}{2}i \) and \( \frac{1}{2} - \frac{\sqrt{5}}{2}i \).

These solutions reflect both the nature of complex numbers and how they provide a meaningful extension to the real number system, allowing us to solve equations that do not have real roots.