Problem 64 - Algebra and Trigonometry

Question

Find all solutions of the equation, and express them in the form \(a+b i\) $$ x^{2}-3 x+3=0 $$

Step-by-Step Solution

Answer

The solutions are \( x_1 = \frac{3}{2} + \frac{\sqrt{3}}{2}i \) and \( x_2 = \frac{3}{2} - \frac{\sqrt{3}}{2}i \).

Step 1: Identify the equation type

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = -3 \), and \( c = 3 \).

Step 2: Calculate the Discriminant

Calculate the discriminant, \( \Delta = b^2 - 4ac \). For our equation, \( \Delta = (-3)^2 - 4(1)(3) = 9 - 12 = -3 \). A negative discriminant indicates that the solutions are complex numbers.

Step 3: Apply the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{\Delta}}{2a} \). Substituting the values \( a = 1 \), \( b = -3 \), and \( \Delta = -3 \), we get \( x = \frac{-(-3) \pm \sqrt{-3}}{2} = \frac{3 \pm \sqrt{-3}}{2} \).

Step 4: Simplify the Square Root of the Negative Discriminant

Simplify \( \sqrt{-3} = i \sqrt{3} \) (since \( \sqrt{-1} = i \)). So \( x = \frac{3 \pm i \sqrt{3}}{2} \).

Step 5: Write the solutions in \( a + bi \) form

The solutions are \( x_1 = \frac{3}{2} + \frac{\sqrt{3}}{2}i \) and \( x_2 = \frac{3}{2} - \frac{\sqrt{3}}{2}i \).

Key Concepts

Complex Numbers

Complex numbers are numbers that consist of a real part and an imaginary part. The imaginary unit is denoted as \( i \), which is defined as the square root of -1: \( i = \sqrt{-1} \). This means that \( i^2 = -1 \). Any complex number can be written in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part.

When it comes to quadratic equations, if the discriminant is negative, the solutions will be complex numbers. For instance, in our given equation, the discriminant is -3. Because it is negative, we know that the solutions will have an imaginary component. The square root of a negative number, like \( \sqrt{-3} \), will always result in an imaginary number. We rewrite this as \( i \sqrt{3} \).

The solutions to the quadratic equation then take the form of complex numbers, precisely combining real and imaginary parts. This reflects how rich the field of mathematics can be, as it expands from the real number line to a 2D plane where each point represents a complex number.

Discriminant

The discriminant is a crucial part of the quadratic equation. Symbolically, it's expressed as \( \Delta = b^2 - 4ac \). The discriminant informs us about the nature of the roots of a quadratic equation:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is one repeated real root, also known as a double root.
• If \( \Delta < 0 \), there are two complex conjugates.

In the equation \( x^2 - 3x + 3 = 0 \), we computed the discriminant as \( (-3)^2 - 4 \times 1 \times 3 = 9 - 12 = -3 \). Since \( \Delta = -3 \), this indicates that the quadratic equation has two complex solutions.

Understanding the discriminant is key to predicting the type of solutions without necessarily solving the entire equation. It gives an overview of what kind of answers to expect, resonating with both real-world applications and more abstract mathematical exploration.

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions of any quadratic equation, even when the coefficients are complex. The formula states:
\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]

This formula applies to any quadratic equation of the form \( ax^2 + bx + c = 0 \). For our specific equation, with \( a = 1 \), \( b = -3 \), and \( c = 3 \), we inserted these values into the formula:

\[ x = \frac{-(-3) \pm \sqrt{-3}}{2 \cdot 1} = \frac{3 \pm \sqrt{-3}}{2} \]

The next step involves simplifying the square root of a negative discriminant as discussed earlier, transforming it to \( i \sqrt{3} \). So, the complete solutions in the standard form \( a + bi \) become:

• \( x_1 = \frac{3}{2} + \frac{\sqrt{3}}{2}i \)
• \( x_2 = \frac{3}{2} - \frac{\sqrt{3}}{2}i \)

The quadratic formula is versatile and extremely useful because it systematically provides the roots of any quadratic equation. It assures you that whether the roots are real or complex, you will find them effectively.