Problem 64 - Precalculus Mathematics for Calculus

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 \(\frac{3}{2} + \frac{i\sqrt{3}}{2}\) and \(\frac{3}{2} - \frac{i\sqrt{3}}{2}\).

Step 1: Identify the quadratic equation

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

Step 2: Calculate the discriminant

The discriminant \(D\) of the quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). Substitute \(a = 1\), \(b = -3\), and \(c = 3\) into the formula:\[ D = (-3)^2 - 4 \times 1 \times 3 \]\[ D = 9 - 12 \]\[ D = -3 \]Since \(D < 0\), the equation has two complex solutions.

Step 3: Use the quadratic formula

The roots of the equation can be found using the quadratic formula:\[ x = \frac{-b \pm \sqrt{D}}{2a} \]Substitute \(a = 1\), \(b = -3\), and \(D = -3\):\[ x = \frac{-(-3) \pm \sqrt{-3}}{2 \times 1} \]\[ x = \frac{3 \pm \sqrt{-3}}{2} \]

Step 4: Simplify the expression for complex roots

Since \(\sqrt{-3} = i\sqrt{3}\), where \(i\) is the imaginary unit, we replace \(\sqrt{-3}\) with \(i\sqrt{3}\) in the expression:\[ x = \frac{3 \pm i\sqrt{3}}{2} \]This results in two complex solutions:\[ x_1 = \frac{3 + i\sqrt{3}}{2} \]\[ x_2 = \frac{3 - i\sqrt{3}}{2} \]

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

Rewrite the above complex numbers in the standard form \(a + bi\):The first solution is:\[ x_1 = \frac{3}{2} + \frac{i\sqrt{3}}{2}\]The second solution is:\[ x_2 = \frac{3}{2} - \frac{i\sqrt{3}}{2}\]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation. A quadratic equation is generally expressed as \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants. The roots of the quadratic equation can be found using the formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] where \( D \) is the discriminant. This formula gives us the solutions to the equation, which may be two distinct real numbers, one real number, or two complex numbers.
In this exercise, the quadratic equation is \( x^2 - 3x + 3 = 0 \). Substituting \( a = 1 \), \( b = -3 \), and \( c = 3 \) into the quadratic formula allows us to solve for \( x \). The quadratic formula helps to quickly determine the solutions without the need for factoring.

Discriminant

The discriminant is a crucial part of the quadratic formula, as it determines the nature of the roots of a quadratic equation. It is calculated using the formula: \[ D = b^2 - 4ac \] Where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation. The value of the discriminant reveals important information about the roots:

• If \( D > 0 \), the equation has two distinct real roots.
• If \( D = 0 \), there is exactly one real root, also known as a double root.
• If \( D < 0 \), the equation has two complex roots.

In our problem, after substituting \( a = 1 \), \( b = -3 \), and \( c = 3 \), we calculate \( D = -3 \). Because \( D < 0 \), we know the equation \( x^2 - 3x + 3 = 0 \) will have two complex solutions. This is an indication that the parabola associated with the quadratic equation does not intersect the x-axis.

Imaginary Unit

When dealing with complex solutions, the imaginary unit \( i \) plays a pivotal role. By definition, the imaginary unit \( i \) is equal to \( \sqrt{-1} \). This means that \( i^2 = -1 \). The imaginary unit is introduced when the discriminant is negative in a quadratic equation.
In the given exercise, the discriminant \( D = -3 \) which means the quadratic formula will yield complex numbers. We see this when we encounter \( \sqrt{-3} \). Instead of halting at a negative square root, we convert it to a form that uses the imaginary unit: \[ \sqrt{-3} = i\sqrt{3} \] Therefore, the solutions \( x = \frac{3 \pm i\sqrt{3}}{2} \) hold the imaginary component. By expressing the solutions in the form \( a + bi \), we clearly define the real part, \( a = \frac{3}{2} \), and the imaginary part, \( b = \frac{\sqrt{3}}{2} \). This shows the crucial role of \( i \) in resolving equations with negative discriminants.