Problem 71 - Precalculus Mathematics for Calculus

Question

Find all solutions of the equation and express them in the form \(a+b i\) $$\frac{1}{2} x^{2}-x+5=0$$

Step-by-Step Solution

Answer

The solutions are \( x = 1 + 3i \) and \( x = 1 - 3i \).

Step 1: Write Down the Quadratic Formula

The general formula for solving a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our equation, \( a = \frac{1}{2} \), \( b = -1 \), and \( c = 5 \).

Step 2: Calculate the Discriminant

The discriminant \( \Delta \) is calculated as \( b^2 - 4ac \). Substitute the given values:\[\Delta = (-1)^2 - 4 \cdot \frac{1}{2} \cdot 5 = 1 - 10 = -9\]Since \( \Delta < 0 \), the solutions will be complex numbers.

Step 3: Use the Quadratic Formula

Plug the values of \( a \), \( b \), and \( \Delta \) into the quadratic formula:\[x = \frac{-(-1) \pm \sqrt{-9}}{2 \cdot \frac{1}{2}}\]Simplifying the expression:\[x = \frac{1 \pm \sqrt{9i^2}}{1}\]

Step 4: Simplify the Complex Numbers

Note that \( \sqrt{-9} = 3i \). Therefore, our expression becomes:\[x = 1 \pm 3i\]Thus, the solutions are \( x = 1 + 3i \) and \( x = 1 - 3i \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of quadratic equations, which are equations that take the form \( ax^2 + bx + c = 0 \). This formula is essential because it provides a method to solve for \( x \) regardless of whether the solutions are real or complex. Here, the quadratic formula is expressed as:

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

This formula requires us to identify the coefficients \( a \), \( b \), and \( c \) from the given quadratic equation.
In our specific exercise equation \( \frac{1}{2}x^2 - x + 5 = 0 \), we determined that:

• \( a = \frac{1}{2} \)
• \( b = -1 \)
• \( c = 5 \)

These values are crucial for our calculations, especially when we move on to compute the discriminant, which is used inside the quadratic formula.

Discriminant

The discriminant is the part of the quadratic formula that helps us determine the nature of the roots of the equation. It is expressed as \( b^2 - 4ac \). The value of the discriminant can reveal whether the solutions will be real or complex:

• If \(\Delta > 0\), the quadratic equation has two distinct real solutions.
• If \(\Delta = 0\), there is exactly one real solution (or a repeated root).
• If \(\Delta < 0\), the solutions are complex or imaginary.

In our calculation:

• \( b = -1 \)
• \( a = \frac{1}{2} \)
• \( c = 5 \)

This gives us:

• \(\Delta = (-1)^2 - 4 \cdot \frac{1}{2} \cdot 5 = 1 - 10 = -9\)

Since \(\Delta < 0\), we know in advance that the roots of the equation will be complex numbers.

Complex Solutions

Complex solutions arise when the discriminant in the quadratic formula is negative. In such cases, the square root of a negative number results in an imaginary unit, \(i\), where \(i^2 = -1\). When using the quadratic formula with a negative discriminant, we encounter solutions involving \(i\).
For our equation, we substitute back into the quadratic formula:

• \[x = \frac{-(-1) \pm \sqrt{-9}}{2 \cdot \frac{1}{2}}\]

Breaking this down:

• The expression simplifies to \[x = \frac{1 \pm \sqrt{9i^2}}{1}\]
• Since \(\sqrt{-9} = 3i\), the resulting solutions are \[x = 1 \pm 3i\]

Thus, the quadratic equation \(\frac{1}{2} x^{2} - x + 5 = 0\) has solutions expressed in the standard complex form \(a + bi\) as \(x = 1 + 3i\) and \(x = 1 - 3i\). Complexity of the solutions is what makes understanding the nature of the discriminant so vital.