Problem 21 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ -3 z^{2}-2 z+4=0 $$

Step-by-Step Solution

Answer

The roots are \(z = \frac{1 + \sqrt{13}}{-3}\) and \(z = \frac{1 - \sqrt{13}}{-3}\).

Step 1: Identify Quadratic Equation Form

The equation given is \(-3z^2 - 2z + 4 = 0\). This is a standard quadratic equation of the form \(ax^2 + bx + c = 0\) with \(a = -3\), \(b = -2\), and \(c = 4\).

Step 2: Calculate the Discriminant

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is calculated as \(\Delta = b^2 - 4ac\). Substituting the given values, we find \(\Delta = (-2)^2 - 4(-3)(4) = 4 + 48 = 52\).

Step 3: Determine Nature of Roots

Since the discriminant \(\Delta = 52\) is greater than zero, the roots of the quadratic equation are real and unequal.

Step 4: Apply the Quadratic Formula

The roots of the equation can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\). Substitute \(a = -3\), \(b = -2\), and \(\Delta = 52\) to get: \(z = \frac{-(-2) \pm \sqrt{52}}{2(-3)} = \frac{2 \pm \sqrt{52}}{-6}\).

Step 5: Simplify the Square Root and Roots

Simplify \(\sqrt{52} = 2\sqrt{13}\). Substitute back: \(z = \frac{2 \pm 2\sqrt{13}}{-6}\). Simplify further: \(z = \frac{1 \pm \sqrt{13}}{-3}\).

Step 6: Find the Final Roots

Thus the roots of the quadratic equation are \(z = \frac{1 + \sqrt{13}}{-3}\) and \(z = \frac{1 - \sqrt{13}}{-3}\).

Step 7: Verify the Solution

Substitute the roots back into the original equation to verify. If both roots satisfy the equation \(-3z^2 - 2z + 4 = 0\), then they are correct. However, this step is optional for understanding since it involves recalculating expressions, which we've solved analytically.

Key Concepts

Discriminant

When solving a quadratic equation, an important concept to understand is the **discriminant**. The discriminant is a specific value that helps you determine the nature of the roots of the quadratic equation. For any quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is given by the formula:\[\Delta = b^2 - 4ac\]This value can tell you a lot about the roots:

• If \( \Delta > 0 \), the quadratic equation has two distinct real roots. This means that the parabola representing the equation crosses the x-axis at two points.
• If \( \Delta = 0 \), there is exactly one real root. This indicates the vertex of the parabola just touches the x-axis, creating a root of multiplicity two.
• If \( \Delta < 0 \), the equation has no real roots. Instead, the roots are complex or imaginary, meaning the parabola does not cross the x-axis at all.

In the exercise, the discriminant for the equation \(-3z^2 - 2z + 4 = 0\) was calculated to be 52. Since 52 is greater than zero, we know that there are two real and distinct roots.

Quadratic Formula

The **Quadratic Formula** is a crucial tool for finding the roots of a quadratic equation, especially when they do not factor easily. The formula is:\[x = \frac{-b \pm \sqrt{\Delta}}{2a}\]To employ this formula, you need the coefficients \(a\), \(b\), and \(c\) from your quadratic equation \(ax^2 + bx + c = 0\), as well as the discriminant \(\Delta = b^2 - 4ac\). The "\(\pm\)" symbol in the formula signifies two potential solutions, capturing the two roots of the equation:

• \(x_1 = \frac{-b + \sqrt{\Delta}}{2a}\)
• \(x_2 = \frac{-b - \sqrt{\Delta}}{2a}\)

In the given problem, we used the Quadratic Formula to solve \(-3z^2 - 2z + 4 = 0\). After calculating the discriminant as 52, the Quadratic Formula gave us the potential roots in the form \(z = \frac{2 \pm \sqrt{52}}{-6}\). Further simplification leads to the real roots presented in the solution.

Real Roots

**Real Roots** of a quadratic equation are the solutions where the equation actually equals zero. These roots can emerge in several scenarios depending on the discriminant value.Because the discriminant in the exercise, 52, is positive, it confirms that the equation \(-3z^2 - 2z + 4 = 0\) has two different real roots. This tells us that the graph of the quadratic equation—a parabola—crosses the x-axis at two unique points. The final roots were calculated as:

• \(z = \frac{1 + \sqrt{13}}{-3}\)
• \(z = \frac{1 - \sqrt{13}}{-3}\)

These values are real numbers and represent the x-coordinates where the parabola representing the quadratic crosses the x-axis. Whenever solving equations, verifying real roots is a good practice to ensure accuracy and a deeper understanding of where a function interacts with the coordinate axes.