Chapter 1: Problem 66
Find all solutions of the equation, and express them in the form \(a+b i\) $$ 2 x^{2}+3=2 x $$
Short Answer
Expert verified
The solutions are \( x = \frac{1}{2} \pm \frac{\sqrt{5}}{2}i \).
Step by step solution
01
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:\[ 2x^2 - 2x + 3 = 0 \]
02
Identify the quadratic formula components
To solve the quadratic equation \( ax^2 + bx + c = 0 \), the coefficients are needed:\[ a = 2, \quad b = -2, \quad c = 3 \]
03
Apply the quadratic formula
The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the coefficients:\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times 3}}{2 \times 2} \]
04
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.
05
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:\[ x = \frac{2 \pm 2\sqrt{5}i}{4} = \frac{1 \pm \sqrt{5}i}{2} \]
06
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 \( a + bi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
The beauty of the quadratic formula is that it works in all cases—whether the solutions are real or complex.
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
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.
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.
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 \( \frac{1}{2} - \frac{\sqrt{5}}{2}i \).