Problem 14 - Algebra for College Students

Question

Solve each quadratic equation by using (a) the factoring method and (b) the method of completing the square. $$n(n-6)=216$$

Step-by-Step Solution

Answer

The solutions for the equation are \( n = 18 \) and \( n = -12 \).

Step 1: Expand and Rearrange the Equation

Begin by expanding the left side of the equation: \( n(n-6) = n^2 - 6n \). Now, rearrange the equation to standard quadratic form: \( n^2 - 6n - 216 = 0 \). This is our equation to solve for \( n \).

Step 2: Solve by Factoring

To factor the quadratic, look for two numbers that multiply to \(-216\) and add to \(-6\). These numbers are \(-18\) and \(12\), so we can factor the quadratic as: \( (n - 18)(n + 12) = 0 \).

Step 3: Find the Roots from Factoring

Set each factor equal to zero to solve for \( n \). From \( n - 18 = 0 \), we have \( n = 18 \). From \( n + 12 = 0 \), we have \( n = -12 \). Thus, the solutions by factoring are \( n = 18 \) and \( n = -12 \).

Step 4: Solve by Completing the Square

Start with the original expanded equation: \( n^2 - 6n - 216 = 0 \). First, move the constant term to the right side: \( n^2 - 6n = 216 \). Then, complete the square by adding \( (-6/2)^2 = 9 \) to both sides: \( n^2 - 6n + 9 = 225 \).

Step 5: Write the Equation as a Perfect Square

The left side is now a perfect square trinomial: \( (n - 3)^2 = 225 \).

Step 6: Solve for n Using the Square Root

Take the square root of both sides: \( n - 3 = \pm 15 \). Now solve for \( n \): \( n = 3 + 15 = 18 \) and \( n = 3 - 15 = -12 \).

Step 7: Verify the Solutions

Both methods give the same solution: \( n = 18 \) and \( n = -12 \). Plug these values back into the original equation to verify correctness.

Key Concepts

Factoring Method

The factoring method is a powerful tool for solving quadratic equations. It involves expressing the equation in the form of a product of two binomials and then setting each binomial factor equal to zero to find the solutions. Let's break it down with an example from the exercise.

The original quadratic equation is given by:

• Expand the left-hand side: \( n(n-6) = n^2 - 6n \).
• Rearrange to standard form: \( n^2 - 6n - 216 = 0 \).

To factor, search for two numbers that multiply to \(-216\) (the constant term). These numbers must also add up to \(-6\) (the coefficient of the linear term).

After considering the possibilities, you find the numbers are \(-18\) and \(12\). Thus, the expression can be rewritten as \( (n - 18)(n + 12) = 0 \).

Setting each factor to zero gives \( n = 18 \) and \( n = -12 \). These are the roots of the quadratic equation. Factoring is often straightforward, but finding the right numbers can sometimes require trial and error or systematic listing of factors.

Completing the Square

Completing the square is another method to solve quadratic equations and is useful when factoring is difficult. This approach involves creating a perfect square trinomial, which simplifies solving for the variable.

Begin with the standard form equation from the exercise:

• \( n^2 - 6n - 216 = 0 \).

First, transfer the constant term to the other side:

• \( n^2 - 6n = 216 \).

Next, add a perfect square to both sides to form a perfect square trinomial. Compute \( (-6/2)^2 = 9 \), and add 9 on both sides to achieve the trinomial:

• \( n^2 - 6n + 9 = 225 \).

Rewrite the left side as a square of a binomial: \( (n - 3)^2 = 225 \).

Now, solve for \( n \) by taking the square root of both sides, giving you \( n - 3 = \pm 15 \). Solve for \( n \) to get the roots: \( n = 18 \) and \( n = -12 \). Completing the square is effective and particularly helpful when graphing or deriving vertex forms.

Quadratic Formula

The quadratic formula provides a foolproof method to solve any quadratic equation, regardless of its factorability. It is derived from the process of completing the square, and it offers a direct way to calculate the roots of the equation.

The formula is given as:

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

This requires identifying the coefficients \(a\), \(b\), and \(c\) from the standard form equation \( an^2 + bn + c = 0 \).

From our example, \( a = 1 \), \( b = -6 \), and \( c = -216 \). Plugging these into the quadratic formula:

• \( n = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-216)}}{2 \cdot 1} \).

After simplifying, you'll find the roots as \( n = 18 \) and \( n = -12 \). This method is particularly useful if the equation is challenging to factor or if you need an exact solution quickly.

Roots of an Equation

The roots of a quadratic equation are the solutions or values of the variable that satisfy the equation. In the context of a parabola, these are the x-intercepts—where the graph crosses the x-axis.

For the quadratic equation \( n^2 - 6n - 216 = 0 \), the roots are \( n = 18 \) and \( n = -12 \). Both solutions can be double-checked by substituting them back into the original equation to ensure correctness.

Understanding roots is crucial because it helps in:

• Graphing the quadratic function accurately by pinpointing the intercepts.
• Solving real-world problems where solutions are desired, such as projectile motion or optimization tasks.

In summary, identifying the roots provides insights into the nature of the quadratic equation and its behavior, contributing significantly to mathematical analysis and applications.