Problem 71 - Algebra and Trigonometry

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: Identify the Quadratic Equation

The given equation is \( \frac{1}{2}x^2 - x + 5 = 0 \). This is a quadratic equation in the general form \( ax^2 + bx + c = 0 \). Here, \( a = \frac{1}{2} \), \( b = -1 \), and \( c = 5 \).

Step 2: Use the Quadratic Formula

The solutions of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 3: Compute the Discriminant

Calculate the discriminant \( b^2 - 4ac \) to determine the nature of the roots. Substitute \( a = \frac{1}{2} \), \( b = -1 \), \( c = 5 \):\[ (-1)^2 - 4 \times \frac{1}{2} \times 5 = 1 - 10 = -9 \]Since the discriminant is negative, the solutions will be complex.

Step 4: Solve Using the Quadratic Formula

Substitute the values \( a = \frac{1}{2} \), \( b = -1 \), and \( \sqrt{b^2 - 4ac} = \sqrt{-9} = 3i \) into the quadratic formula:\[ x = \frac{-(-1) \pm 3i}{2 \times \frac{1}{2}} = \frac{1 \pm 3i}{1} = 1 \pm 3i \]

Step 5: Express Solutions in Standard Form

The solutions \( x = 1 + 3i \) and \( x = 1 - 3i \) are in the standard form \( a + bi \), where \( a = 1 \) and \( b = \pm 3 \).

Key Concepts

Complex Numbers

Complex numbers are an extension of the real numbers and are used to solve equations that do not have real solutions. A complex number is expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit defined by \(i^2 = -1\). This means any number in this form has a real part \(a\) and an imaginary part \(b\).

When dealing with quadratic equations, complex numbers may emerge if the discriminant is negative, indicating no real solutions. This is what happens in the equation \(\frac{1}{2}x^2 - x + 5 = 0\). The solutions we found, \(1 + 3i\) and \(1 - 3i\), are examples of complex numbers.

• The real part of the solutions is 1.
• The imaginary part is \(\pm 3i\).

Knowing how to work with complex numbers allows us to solve quadratic equations that would otherwise seem impossible to tackle with only real numbers.

Discriminant

The discriminant is a key component of the quadratic formula, and it helps us determine the nature of the roots of a quadratic equation. The discriminant is represented by \(b^2 - 4ac\) in the quadratic formula, where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation \(ax^2 + bx + c = 0\).

The general guidelines for interpreting the discriminant are:

• If \(b^2 - 4ac > 0\), the quadratic equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), there is one real root, sometimes called a repeated or double root.
• If \(b^2 - 4ac < 0\), the equation has no real roots and two complex conjugate roots instead.

In our specific exercise, calculating \((-1)^2 - 4 \times \frac{1}{2} \times 5 = -9\) yielded a negative discriminant, leading to complex solutions. Understanding the nature of the discriminant helps predict not only whether solutions will exist but also what type they will be, allowing for more tailored approaches when solving quadratic equations.

Quadratic Formula

The quadratic formula is a powerful tool in solving quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

This formula can help find the roots of any quadratic equation by merely plugging in the coefficient values \(a\), \(b\), and \(c\). The expression under the square root, \(b^2 - 4ac\), is known as the discriminant.

• If the discriminant is positive, we find two real and distinct solutions.
• If zero, there’s exactly one real solution.
• If negative, the results are complex solutions.

By substituting \(a = \frac{1}{2}\), \(b = -1\), and \(c = 5\), and computing the discriminant \(b^2 - 4ac = -9\), the quadratic formula led us to the complex roots \(1 + 3i\) and \(1 - 3i\). This is why the quadratic formula is so versatile—it handles both real and complex solutions with ease.