Problem 50 - Intermediate Algebra

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 to the equation are \(n = \frac{1 + \sqrt{3}}{2}\) and \(n = \frac{1 - \sqrt{3}}{2}\).

Step 1: Identify the Equation Type

Recognize that the equation \(2n^{2} - 2n - 1 = 0\) is a quadratic equation because it is in the form \(ax^{2} + bx + c = 0\) where \(a = 2\), \(b = -2\), and \(c = -1\).

Step 2: Check for Simple Factoring

Attempt to factor the equation by finding two numbers that multiply to \(2 \times -1 = -2\) and add to \(-2\). No such integer pair exists, so simple factoring is not viable.

Step 3: Use the Quadratic Formula

Use the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(n\). Here, \(a = 2\), \(b = -2\), and \(c = -1\).

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\). Substitute the values: \((-2)^2 - 4 \times 2 \times (-1) = 4 + 8 = 12\). The positive discriminant indicates two real and distinct solutions.

Step 5: Solve for Roots

Substitute the discriminant back into the quadratic formula: \[n = \frac{-(-2) \pm \sqrt{12}}{2 \times 2}\] which simplifies to \[n = \frac{2 \pm \sqrt{12}}{4}\].

Step 6: Simplify the Square Root

Simplify \(\sqrt{12}\) to \(2\sqrt{3}\) so the formula becomes \[n = \frac{2 \pm 2\sqrt{3}}{4}\].

Step 7: Simplify the Equation

Divide each term by 2: \[n = \frac{1 \pm \sqrt{3}}{2}\]. Thus, the solutions are \(n = \frac{1 + \sqrt{3}}{2}\) and \(n = \frac{1 - \sqrt{3}}{2}\).

Key Concepts

Quadratic Formula

In the realm of algebra, quadratic equations are quite significant. They appear in the standard form \(ax^2 + bx + c = 0\). To find solutions to these equations, one very powerful tool is the quadratic formula. This formula is expressed as:

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

Here, \(a\), \(b\), and \(c\) come from the coefficients of the quadratic equation. Using this formula, you can directly find the roots (solutions) of any quadratic equation. It's particularly useful when the equation doesn't easily factorize. Remember:

• "\(\pm\)" indicates there are generally two solutions.
• Solving involves basic operations of addition, subtraction, multiplication, and division.

This versatility makes the quadratic formula an indispensable part of solving quadratic equations.

Discriminant

The discriminant is a crucial part of the quadratic formula and significantly influences the nature of the roots. Calculated as \(b^2 - 4ac\), the discriminant provides key insights:

• If it's positive, there are two distinct real solutions.
• If it's zero, there's exactly one real solution, meaning the solutions are equal. This results in double roots.
• If it's negative, the equation has two complex solutions since the square root of a negative number emerges.

For the equation \(2n^2 - 2n - 1 = 0\), the discriminant is computed as follows:

• \((-2)^2 - 4 \cdot 2 \cdot (-1) = 4 + 8 = 12\)

A positive discriminant of 12 indicates two distinct real solutions. Understanding the discriminant helps predict how many and what type of solutions to expect before diving deeper.

Factoring

Factoring is another method to solve quadratic equations but isn't always applicable. The idea is to transform the equation into a product of binomials. For an equation like \(ax^2 + bx + c = 0\), you look for two numbers that multiply to \(ac\) (the product of the leading coefficient \(a\) and the constant term \(c\)) and add to \(b\), the middle term.

• Eg: Find numbers that multiply to \(2 \cdot -1 = -2\) and add to \(-2\).
• If such pairs can't be easily determined or don't exist, factoring may not be viable.

For our equation, no integer solution pairs meet the criteria, implying that factoring is impractical. When factoring fails, the quadratic formula becomes particularly useful as it guarantees solutions.

Square Root Simplification

Simplifying square roots is a crucial step, especially when using the quadratic formula. Once you have determined the expression within the square root (the discriminant, \(b^2 - 4ac\)), simplifying it can reveal more straightforward results. For example, with a discriminant of 12, the square root simplifies:

• \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\)

Simplifying square roots involves factoring out perfect squares (like 4, 9, 16, etc.). These make calculations cleaner and results more interpretable.
In the original problem, square root simplification turned the quadratic formula result into an easier-to-manage form, which was then further simplified to find the distinct values for \(n\). This step is key to achieving the most elegant—and correct—version of your solution.