Solving Quadratic Equations Finding Zeros And Fractional Solutions

To effectively find the zeros of the quadratic function f(x) = 9x² + 6x + 1, we will delve into a step-by-step process using the quadratic formula. This method is particularly useful for quadratic equations that are not easily factorable. Understanding how to apply the quadratic formula is crucial in algebra and calculus, providing a reliable way to determine the roots or solutions of any quadratic equation. In this article, we will not only solve the given equation but also provide a comprehensive explanation of the underlying concepts and why this approach is effective.

Understanding the Quadratic Formula

The quadratic formula is derived from the standard form of a quadratic equation, which is ax² + bx + c = 0. Here, a, b, and c are coefficients, with a not equal to zero. The quadratic formula itself is expressed as:

x = (-b ± √(b² - 4ac)) / (2a)

This formula provides two possible solutions for x, represented by the ± symbol, which indicates that we need to calculate the result for both the addition and subtraction cases. The expression inside the square root, b² - 4ac, is known as the discriminant. The discriminant plays a crucial role in determining the nature of the roots:

• If b² - 4ac > 0, the equation has two distinct real roots.
• If b² - 4ac = 0, the equation has one real root (a repeated root).
• If b² - 4ac < 0, the equation has two complex roots.

Understanding the discriminant helps in predicting the type of solutions we should expect before fully applying the formula. This can save time and prevent errors in calculation.

Applying the Quadratic Formula to Our Equation

For our specific equation, f(x) = 9x² + 6x + 1, we identify the coefficients as follows:

• a = 9
• b = 6
• c = 1

Now, we substitute these values into the quadratic formula:

x = (-6 ± √(6² - 4 * 9 * 1)) / (2 * 9)

This substitution is the first concrete step in solving for the zeros. It's crucial to ensure that the values are correctly placed to avoid computational errors. The next step involves simplifying the expression to reveal the solutions.

Step-by-Step Solution

Now, let's simplify the expression we obtained after substituting the values into the quadratic formula:

x = (-6 ± √(6² - 4 * 9 * 1)) / (2 * 9)

1. Simplify the Discriminant

The first part of the simplification involves calculating the discriminant, which is the expression under the square root:

√(6² - 4 * 9 * 1) = √(36 - 36) = √0

As we can see, the discriminant equals 0. This indicates that the quadratic equation has exactly one real root (a repeated root). This is a significant observation because it tells us what kind of solution to expect.

2. Substitute the Discriminant Value

Now, we substitute the value of the discriminant back into the quadratic formula:

x = (-6 ± √0) / (2 * 9)

Since the square root of 0 is 0, the equation simplifies further:

x = (-6 ± 0) / (2 * 9)

3. Simplify the Denominator

Next, we simplify the denominator:

2 * 9 = 18

So, the equation now looks like this:

x = -6 / 18

4. Reduce the Fraction

Finally, we reduce the fraction to its simplest form. Both the numerator and the denominator are divisible by 6:

x = -6 / 18 = -1 / 3

Therefore, the quadratic equation f(x) = 9x² + 6x + 1 has one real root, which is x = -1/3. This result demonstrates the power and efficiency of the quadratic formula in solving quadratic equations.

Verifying the Solution

To ensure the accuracy of our solution, we can substitute x = -1/3 back into the original quadratic equation f(x) = 9x² + 6x + 1 and check if it equals zero:

f(-1/3) = 9(-1/3)² + 6(-1/3) + 1

1. Calculate the Square

First, calculate the square of -1/3:

(-1/3)² = 1/9

2. Substitute and Simplify

Now, substitute this back into the equation:

f(-1/3) = 9(1/9) + 6(-1/3) + 1

Simplify each term:

9(1/9) = 1
6(-1/3) = -2

So, the equation becomes:

f(-1/3) = 1 - 2 + 1

3. Final Calculation

Perform the final calculation:

1 - 2 + 1 = 0

Since f(-1/3) = 0, our solution x = -1/3 is correct. This verification step is crucial in mathematical problem-solving as it confirms the accuracy of the result and understanding of the process.

Alternative Methods: Factoring

While the quadratic formula is a universal method for solving quadratic equations, some equations can be solved more efficiently through factoring. Factoring involves expressing the quadratic equation as a product of two binomials. Let's see if we can factor our equation, f(x) = 9x² + 6x + 1.

The given quadratic equation is:

9x² + 6x + 1

1. Recognize the Pattern

Notice that the equation can be seen as a perfect square trinomial. A perfect square trinomial is of the form a² + 2ab + b² or a² - 2ab + b², which can be factored into (a + b)² or (a - b)², respectively. In our case:

• 9x² is the square of 3x
• 1 is the square of 1
• 6x is twice the product of 3x and 1 (i.e., 2 * 3x * 1 = 6x)

2. Factor the Trinomial

Thus, we can rewrite the equation as:

(3x + 1)²

3. Solve for x

To find the zeros, we set the factored form equal to zero:

(3x + 1)² = 0

Taking the square root of both sides:

3x + 1 = 0

Solving for x:

3x = -1
x = -1/3

As we can see, factoring provides the same solution, x = -1/3, as the quadratic formula. Factoring, when applicable, can be a quicker method, but it requires recognizing specific patterns in the equation. The quadratic formula, on the other hand, is a reliable method that works for all quadratic equations.

The Discriminant and Its Role

The discriminant, represented as b² - 4ac, is a critical component of the quadratic formula that provides valuable information about the nature of the roots of a quadratic equation without fully solving the equation. Understanding the discriminant helps in predicting whether the roots will be real and distinct, real and equal, or complex.

1. Positive Discriminant (b² - 4ac > 0)

If the discriminant is positive, the quadratic equation has two distinct real roots. This means the parabola intersects the x-axis at two different points. For example, if we had an equation where the discriminant calculated to be 16, then we would have two different real solutions because √16 is a real number (4).

2. Zero Discriminant (b² - 4ac = 0)

If the discriminant is zero, the quadratic equation has exactly one real root, which is a repeated root. This means the vertex of the parabola touches the x-axis at one point. In our original equation, 9x² + 6x + 1, the discriminant was 0, leading to the single solution x = -1/3.

3. Negative Discriminant (b² - 4ac < 0)

If the discriminant is negative, the quadratic equation has two complex roots. This means the parabola does not intersect the x-axis. For instance, if the discriminant was -16, we would have complex solutions because we would need to take the square root of -16, which is not a real number.

Understanding and calculating the discriminant as an initial step when solving quadratic equations can save time and provide insight into the nature of the solutions. It's a fundamental concept in algebra that helps in various mathematical applications.

Real-World Applications of Quadratic Equations

Quadratic equations are not just abstract mathematical concepts; they have numerous applications in real-world scenarios. Their ability to model curved paths and relationships makes them indispensable in various fields, ranging from physics and engineering to economics and computer science. Let's explore some practical examples where quadratic equations are used.

1. Physics: Projectile Motion

One of the most common applications of quadratic equations is in describing projectile motion. When an object is thrown or launched into the air, its path follows a parabolic trajectory, which can be modeled by a quadratic equation. The equation takes into account factors such as initial velocity, launch angle, and gravitational acceleration. For example, engineers use quadratic equations to calculate the trajectory of a ball thrown, a rocket launched, or the distance a projectile will travel.

The general form of the equation for the height h of a projectile as a function of time t is:

h(t) = -1/2gt² + v₀t + h₀

Where:

• g is the acceleration due to gravity (approximately 9.8 m/s² on Earth)
• v₀ is the initial vertical velocity
• h₀ is the initial height

This equation helps in determining the maximum height reached by the projectile, the time it takes to reach the ground, and the horizontal range.

2. Engineering: Bridge Design

Engineers use quadratic equations in the design of bridges, particularly suspension bridges and arches. The parabolic shape of suspension cables and arches can be modeled using quadratic functions. By understanding the properties of parabolas, engineers can ensure that bridges are structurally sound and can withstand various loads. The equations help calculate the tension in the cables, the curvature of the arch, and the overall stability of the structure.

3. Economics: Cost and Revenue Optimization

In economics, quadratic functions are used to model cost, revenue, and profit functions. For example, the cost of production might increase quadratically with the quantity produced due to factors like overtime pay and diminishing returns. Similarly, the revenue might follow a quadratic pattern as price affects demand. Businesses can use these quadratic models to determine the optimal production level or pricing strategy that maximizes profit. The vertex of the parabola in these models often represents the point of maximum profit or minimum cost.

4. Computer Graphics: Curve Modeling

Quadratic equations play a crucial role in computer graphics for modeling curves and surfaces. Bezier curves, which are widely used in graphic design and animation, are often defined using quadratic or cubic equations. These curves allow designers to create smooth, scalable shapes and paths. From designing fonts to creating animated characters, quadratic equations are fundamental in generating visual elements on computers.

5. Optics: Lenses and Mirrors

The shape of lenses and mirrors in optical devices, such as telescopes and cameras, can be described using quadratic equations. The curvature of a lens determines how it focuses light, and quadratic equations help in designing lenses that produce clear and focused images. The parabolic shape of mirrors in telescopes, for instance, is crucial for focusing light from distant stars onto a single point.

These examples illustrate the broad applicability of quadratic equations in various fields. Their ability to model real-world phenomena accurately makes them an essential tool for scientists, engineers, economists, and computer professionals.

Conclusion

In conclusion, we have effectively solved for the zeros of the quadratic function f(x) = 9x² + 6x + 1 using the quadratic formula, resulting in a single real root x = -1/3. We also verified this solution by substituting it back into the original equation. Furthermore, we explored an alternative method of factoring, which confirmed our result, and discussed the significance of the discriminant in determining the nature of the roots. Finally, we highlighted the diverse real-world applications of quadratic equations across various fields.

Mastering the use of the quadratic formula and understanding the concepts related to quadratic equations are essential skills in mathematics. The ability to solve these equations not only enhances problem-solving capabilities but also provides a foundation for more advanced mathematical studies and practical applications. Whether in physics, engineering, economics, or computer science, quadratic equations offer a powerful tool for modeling and understanding the world around us. By practicing and applying these techniques, one can gain a deeper appreciation for the versatility and importance of quadratic equations.