Problem 7 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ 2 z^{2}=13 z+15 $$

Step-by-Step Solution

Answer

The solutions are \( z = 7.5 \) and \( z = -1 \).

Step 1: Write the Equation in Standard Form

Start by bringing all terms to one side of the equation. The given equation is \( 2z^2 = 13z + 15 \). Subtract \( 13z \) and \( 15 \) from both sides to get: \[ 2z^2 - 13z - 15 = 0 \] This is the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \).

Step 2: Identify Coefficients

Identify the coefficients from the quadratic equation \( 2z^2 - 13z - 15 = 0 \). Here, \( a = 2 \), \( b = -13 \), and \( c = -15 \).

Step 3: Use the Quadratic Formula

The quadratic formula is given by \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the values of \( a \), \( b \), and \( c \) into the formula: \[ z = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2} \]

Step 4: Calculate the Discriminant

First, calculate the discriminant \( b^2 - 4ac \): \[ (-13)^2 - 4 \cdot 2 \cdot (-15) = 169 + 120 = 289 \]

Step 5: Solve for Roots

Now that we have the discriminant, substitute it back into the quadratic formula: \[ z = \frac{13 \pm \sqrt{289}}{4} \] Since \( \sqrt{289} = 17 \), we have: \[ z = \frac{13 \pm 17}{4} \] Thus, the solutions are: \[ z_1 = \frac{13 + 17}{4} = 7.5 \] \[ z_2 = \frac{13 - 17}{4} = -1 \]

Step 6: Verify the Solutions

To verify, substitute \( z_1 = 7.5 \) and \( z_2 = -1 \) back into the original equation: For \( z_1 = 7.5 \):\[ 2(7.5)^2 = 13(7.5) + 15 \] \[ 112.5 = 112.5 \] For \( z_2 = -1 \): \[ 2(-1)^2 = 13(-1) + 15 \] \[ 2 = 2 \] Both solutions satisfy the original equation.

Key Concepts

Quadratic Formula

When you have a quadratic equation in the form of \( ax^2 + bx + c = 0 \), you can always rely on the quadratic formula to find the roots, or solutions, of the equation.
The quadratic formula is expressed as:

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

This nifty tool allows us to solve any quadratic equation, no matter how complex. Using the values for \( a \), \( b \), and \( c \) from your specific problem, you can substitute directly into the formula to find the possible values of \( x \).
For example, in our given exercise, where \( a = 2 \), \( b = -13 \), and \( c = -15 \), you only need to plug these into the formula to get the solutions:

• \( z = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2} \)

The quadratic formula notably includes a portion under the square root, known as the discriminant. This plays a key role in determining the nature of the roots, which we'll discuss further in later sections.

Standard Form of Quadratic Equation

To effectively apply the quadratic formula, your equation must first be in the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \).
In this form, \( a \), \( b \), and \( c \) represent numerical constants. This format is crucial for executing various solving techniques, including use of the quadratic formula.
Consider the given problem: originally \( 2z^2 = 13z + 15 \).
You'll want to bring everything to one side to convert it to standard form:

• Subtract \( 13z \) and \( 15 \) from both sides resulting in \( 2z^2 - 13z - 15 = 0 \).
• Here, \( a = 2 \), \( b = -13 \), and \( c = -15 \).

Once in this form, you can clearly see what values you'll be working with, making it much easier to move forward with solving the equation.

Discriminant in Quadratic Equations

The discriminant is a component of the quadratic formula, appearing under the square root as \( b^2 - 4ac \).
It tells us more than just the number of solutions; it describes the nature of the solutions as well:

• If the discriminant is positive, there are two real and distinct solutions.
• If it's zero, there's exactly one real solution.
• If negative, there are two complex (imaginary) solutions.

In the example problem, \( b = -13 \), \( a = 2 \), and \( c = -15 \), thus the discriminant is:

• \[ (-13)^2 - 4 \cdot 2 \cdot (-15) = 289 \]
• Since \( 289 \) is positive and a perfect square, the roots are real and rational: \( 7.5 \) and \( -1 \).

Understanding the discriminant helps you anticipate the kinds of solutions you'll find, allowing you to predict the characteristics of the equation's roots even before finding the exact values.