Problem 121 - College Algebra with Modeling and Visualization

Question

Discuss three symbolic methods for solving a quadratic equation. Make up a quadratic equation and use each method to find the solution set.

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(x^2 - 5x + 6 = 0\) are \(x = 2\) and \(x = 3\) using factoring, completing the square, and the quadratic formula.

Step 1: Identify and Write a Quadratic Equation

A quadratic equation is generally in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Let's consider the quadratic equation \( x^2 - 5x + 6 = 0 \). This equation is chosen as an example containing real coefficients.

Step 2: Solving by Factoring

To solve the equation \( x^2 - 5x + 6 = 0 \) by factoring, we look for two numbers that multiply to 6 (the constant term) and add to -5 (the linear coefficient). These numbers are -2 and -3, so the equation factors as \( (x - 2)(x - 3) = 0 \). Setting each factor equal to zero gives us the solutions \( x = 2 \) and \( x = 3 \).

Step 3: Solving by Completing the Square

For the equation \( x^2 - 5x + 6 = 0 \), we first move 6 to the other side: \( x^2 - 5x = -6 \). Next, add \( (\frac{-5}{2})^2 = \frac{25}{4} \) to both sides: \( x^2 - 5x + \frac{25}{4} = \frac{25}{4} - 6 \). Simplify to find \( x^2 - 5x + \frac{25}{4} = \frac{1}{4} \). This is equivalent to \( (x - \frac{5}{2})^2 = \frac{1}{4} \). Solving gives \( x - \frac{5}{2} = \pm \frac{1}{2} \). Hence, the solutions are \( x = 3 \) and \( x = 2 \).

Step 4: Solving using the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For the equation \( x^2 - 5x + 6 = 0 \), identify \( a = 1 \), \( b = -5 \), and \( c = 6 \). Substitute these into the formula: \( x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \). Simplify to get \( x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} \). Therefore, the solutions are \( x = 3 \) and \( x = 2 \).

Key Concepts

Factoring Method

The factoring method for solving quadratic equations is often a favored approach because of its simplicity and speed. It involves expressing the quadratic equation in the form \[ (x - p)(x - q) = 0 \]where \( p \) and \( q \) are numbers that satisfy both of the following conditions:

• They multiply to give you the constant term \( c \) at the end of the equation.
• They add up to give you the middle coefficient \( b \).

Once you have the equation written this way, finding the solutions becomes straightforward because you can set each factor equal to zero. For instance, with an equation like \[ x^2 - 5x + 6 = 0 \]we look for two numbers that multiply to 6 and add up to -5. The numbers \(-2\) and \(-3\) work since \((-2) \times (-3) = 6\) and \(-2) + (-3) = -5\). You then solve the derived linear equations:\( x - 2 = 0 \) which gives \( x = 2 \), and \( x - 3 = 0 \) which gives \( x = 3 \).
The solutions \( x = 2 \) and \( x = 3 \) are the roots of the quadratic equation, meaning they are the values where the equation equals zero.

Completing the Square

Completing the square is a systematic way to solve a quadratic equation by turning it into a perfect square trinomial. The goal is to manipulate the original equation to a form like:\[ (x - d)^2 = e \]Let's see this in action with our example equation \( x^2 - 5x + 6 = 0 \).

• First, move the constant term to the right: \( x^2 - 5x = -6 \).
• To complete the square, add the square of half the coefficient of \( x \), which in this case is \( \left(\frac{-5}{2}\right)^2 = \frac{25}{4} \), to both sides.
• You now have \( x^2 - 5x + \frac{25}{4} = \frac{1}{4} \). The left side is a perfect square trinomial \((x - \frac{5}{2})^2 \).

From \( (x - \frac{5}{2})^2 = \frac{1}{4} \), solve for \( x \) using the property \( a^2 = b\) which gives \( a = \pm \sqrt{b} \). So you get:

• \( x - \frac{5}{2} = \pm \frac{1}{2} \)
• which yields two solutions: \( x = 3 \) and \( x = 2 \).

This method is particularly useful when the quadratic cannot be easily factored, and it forms the basis for deriving the quadratic formula.

Quadratic Formula

The quadratic formula is a foolproof method for solving any quadratic equation \[ ax^2 + bx + c = 0 \]It is derived by completing the square in the general case and provides the solutions directly using the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this application, identify the coefficients \( a \), \( b \), and \( c \) from your equation. Let's look at \( x^2 - 5x + 6 = 0 \)where \( a = 1 \), \( b = -5 \), and \( c = 6 \). Substitute these values into the formula:

• Calculate the discriminant \( b^2 - 4ac \), which informs the nature of the roots. Here, it's \( 25 - 24 = 1 \).
• Plug everything into the formula: \[ x = \frac{-(-5) \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2} \]

This simplifies to two possible solutions: \( x = \frac{6}{2} = 3 \) and \( x = \frac{4}{2} = 2 \).
The quadratic formula is extremely powerful, offering a solution pathway regardless of complex numbers or irrational roots. It guarantees solving any quadratic equation when other methods may be tedious or impossible.