Problem 50 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate. $$2 n^{2}-2 n-1=0$$

Step-by-Step Solution

Answer

The solutions are \(n = \frac{1}{2} + \frac{\sqrt{3}}{2}\) and \(n = \frac{1}{2} - \frac{\sqrt{3}}{2}\).

Step 1: Identify the Quadratic Equation Form

The given equation is \[2n^2 - 2n - 1 = 0.\]It is a quadratic equation because it is of the form \[ax^2 + bx + c = 0\]where \(a = 2\), \(b = -2\), and \(c = -1\).

Step 2: Consider Appropriate Methods

To solve this quadratic equation, we could use factoring, completing the square, or the quadratic formula. Since the equation does not easily factor, the quadratic formula is a suitable choice.

Step 3: Apply the Quadratic Formula

The quadratic formula is given by \[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]Substitute \(a = 2\), \(b = -2\), and \(c = -1\) into the formula.

Step 4: Calculate the Discriminant

First, calculate the discriminant, \(b^2 - 4ac\):\[(-2)^2 - 4 \times 2 \times (-1) = 4 + 8 = 12.\]

Step 5: Substitute Values Into Formula

Substituting the values into the quadratic formula gives:\[n = \frac{-(-2) \pm \sqrt{12}}{2 \times 2}.\]Simplify to:\[n = \frac{2 \pm \sqrt{12}}{4}.\]

Step 6: Simplify the Expressions

Simplify \(\sqrt{12}\):\[\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}.\]Thus, \[n = \frac{2 \pm 2\sqrt{3}}{4} = \frac{2}{4} \pm \frac{2\sqrt{3}}{4} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}.\]

Step 7: Write the Final Solution

The solutions to the equation are:\[n = \frac{1}{2} + \frac{\sqrt{3}}{2}\] and \[n = \frac{1}{2} - \frac{\sqrt{3}}{2}.\]

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). This formula can help find the solutions, or roots, of these equations when they are difficult to factor. The quadratic formula is expressed as:

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

Here, \(a\), \(b\), and \(c\) are the coefficients of the terms in the quadratic equation. This formula provides two solutions because of the 'plus-minus' (\(\pm\)) sign, which indicates there are typically two roots for a quadratic equation. The solutions are derived by plugging the quadratic equation's coefficients into the formula. It is essential in various mathematical and practical applications, providing a straightforward approach to finding unknown variables.
In our example, where the equation is \(2n^2 - 2n - 1 = 0\), substituting \(a = 2\), \(b = -2\), and \(c = -1\) into the formula allows us to determine the values of \(n\) that satisfy the equation.

Discriminant

The discriminant is part of the quadratic formula and is critical in determining the nature of the solutions to a quadratic equation. It is the component under the square root, \(b^2 - 4ac\), within the quadratic formula.

• If the discriminant is positive, \(b^2 - 4ac > 0\), there are two distinct real roots.
• If it is zero, \(b^2 - 4ac = 0\), there is exactly one real root, also known as a repeated or double root.
• If the discriminant is negative, \(b^2 - 4ac < 0\), then there are no real roots; instead, there are two complex roots.

In our specific equation \(2n^2 - 2n - 1 = 0\), we calculated the discriminant as \(12\), which is positive. This indicates that the equation has two distinct real roots.
The discriminant gives us a quick insight into what type of numbers we will get from the roots and the nature of the solutions even before solving the full equation.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in algebra, involving various methods like factoring, completing the square, or using the quadratic formula. The goal is to find the roots of the equation, which are the values of the variable that make the equation true.

• Factoring involves expressing the quadratic equation as a product of two binomials. However, this method can be tricky if the equation does not easily factor.
• Completing the square involves rearranging terms to create a perfect square trinomial, which can then be solved by taking the square root.
• The quadratic formula method is often more straightforward, especially if the equation is complex, as it systematically derives the solutions using the coefficients.

In our example, we applied the quadratic formula to solve \(2n^2 - 2n - 1 = 0\) as it does not factor neatly. By using the formula, we quickly found the roots \(n = \frac{1}{2} + \frac{\sqrt{3}}{2}\) and \(n = \frac{1}{2} - \frac{\sqrt{3}}{2}\).
This approach underscores the formula's versatility and reliability in solving any quadratic equation.