Problem 56 - Intermediate Algebra

Question

For each quadratic equation, first use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions. Then solve the equation. $$ 2 x^{2}-x+5=0 $$

Step-by-Step Solution

Answer

Two nonreal complex solutions: \(x = \frac{1 \pm i\sqrt{39}}{4}\).

Step 1: Identify the coefficients

The given quadratic equation is \(2x^2 - x + 5 = 0\). By comparing it with the standard form \(ax^2 + bx + c = 0\), we identify the coefficients: \(a = 2\), \(b = -1\), and \(c = 5\).

Step 2: Calculate the discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\) into the formula: \(D = (-1)^2 - 4 \times 2 \times 5\). This simplifies to \(D = 1 - 40 = -39\).

Step 3: Interpret the discriminant

Since the discriminant \(D = -39\) is less than zero, the quadratic equation has two nonreal complex solutions.

Step 4: Solve the equation using the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{D}}{2a}\). Substitute \(a = 2\), \(b = -1\), and \(D = -39\) into the formula: \(x = \frac{-(-1) \pm \sqrt{-39}}{2 \times 2}\). This becomes \(x = \frac{1 \pm i\sqrt{39}}{4}\).

Step 5: Present the solutions

The solutions to the quadratic equation \(2x^2 - x + 5 = 0\) are \(x = \frac{1 + i\sqrt{39}}{4}\) and \(x = \frac{1 - i\sqrt{39}}{4}\), which are nonreal complex numbers.

Key Concepts

Discriminant

The discriminant is a critical component of quadratic equations. It helps determine the nature of the roots without actually solving the equation. For a quadratic equation in the standard form of \( ax^2 + bx + c = 0 \), the discriminant is given by the formula \( D = b^2 - 4ac \). Here is what the discriminant tells us:

• If \( D > 0 \), the equation has two distinct real roots.
• If \( D = 0 \), it has one real root with a multiplicity of two (or a double root).
• If \( D < 0 \), there are two complex roots (not real).

In our example, the discriminant is \( -39 \). As this is less than zero, the equation has two nonreal complex solutions. Understanding the discriminant not only saves time, but also gives insight into the properties of the solutions before computation.

Complex Solutions

When dealing with non-positive discriminants like \( D < 0 \), we encounter complex solutions in quadratic equations. Complex numbers include a real part and an imaginary part. They are expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit \( \sqrt{-1} \).

For example, in the quadratic equation \( 2x^2 - x + 5 = 0 \), the discriminant was negative, indicating the solutions are complex numbers. When we applied the quadratic formula, we obtained solutions of the form \( \frac{1 \pm i\sqrt{39}}{4} \). Each solution includes an imaginary part (\( i\sqrt{39} \)) due to the negative discriminant. Understanding imaginary components is essential in complex number operations and in interpreting solutions that are not real.

Quadratic Formula

The quadratic formula provides a method to find the solutions of any quadratic equation. It is a cornerstone of algebra and makes solving complex equations straightforward. The formula is:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula uses the coefficients from the equation \( ax^2 + bx + c = 0 \). The term \( \sqrt{b^2 - 4ac} \) is the discriminant, determining the type of solutions. The '±' symbol indicates two potential solutions — often the roots of the equation.

In our problem, substituting \( a = 2 \), \( b = -1 \), and \( c = 5 \) into the formula, we find the solutions: \( \frac{1 \pm i\sqrt{39}}{4} \). These computations demonstrate how the quadratic formula efficiently handles complex numbers through its inclusion of the discriminant. Therefore, mastering the quadratic formula is essential for addressing quadratic equations effectively.