Solving V² - 9v + 14 = 0 A Step-by-Step Guide

In the realm of mathematics, quadratic equations hold a prominent position. These equations, characterized by the presence of a squared term, often appear in various fields, from physics and engineering to economics and computer science. Mastering the techniques for solving quadratic equations is thus an essential skill for any aspiring mathematician or scientist.

Understanding Quadratic Equations

At its core, a quadratic equation is a polynomial equation of the second degree. Its general form is expressed as:

ax² + bx + c = 0

where 'a', 'b', and 'c' are constants, and 'x' represents the variable we aim to solve for. The solutions to a quadratic equation, also known as its roots or zeros, are the values of 'x' that satisfy the equation. These roots represent the points where the parabola defined by the quadratic equation intersects the x-axis.

In this comprehensive guide, we will delve into the methods for solving quadratic equations, focusing on the specific example: v² - 9v + 14 = 0. By the end of this article, you will be well-equipped to tackle similar equations and gain a deeper understanding of quadratic equations.

Methods for Solving Quadratic Equations

There are several techniques available for solving quadratic equations, each with its own strengths and weaknesses. The most common methods include:

1. Factoring: This method involves expressing the quadratic equation as a product of two linear factors. It is a relatively straightforward approach when the equation can be easily factored.
2. Completing the Square: This technique involves manipulating the equation to create a perfect square trinomial on one side, allowing us to isolate the variable and solve for its values.
3. Quadratic Formula: This formula provides a direct solution for any quadratic equation, regardless of its factorability. It is a versatile method that guarantees a solution, even when factoring proves difficult.

Solving v² - 9v + 14 = 0 by Factoring

Our given equation, v² - 9v + 14 = 0, lends itself well to the factoring method. The goal here is to rewrite the quadratic expression as a product of two binomials. To achieve this, we seek two numbers that:

• Multiply to give the constant term (14)
• Add up to the coefficient of the linear term (-9)

After some contemplation, we identify the numbers -2 and -7 as the ideal candidates. Their product is (-2) * (-7) = 14, and their sum is (-2) + (-7) = -9. Thus, we can rewrite the equation as:

(v - 2)(v - 7) = 0

Now, the principle of zero products comes into play. This principle states that if the product of two factors is zero, then at least one of the factors must be zero. Applying this to our factored equation, we have two possibilities:

1. v - 2 = 0, which implies v = 2
2. v - 7 = 0, which implies v = 7

Therefore, the solutions to the quadratic equation v² - 9v + 14 = 0 are v = 2 and v = 7.

Verifying the Solutions

To ensure the accuracy of our solutions, we can substitute them back into the original equation. Let's start with v = 2:

(2)² - 9(2) + 14 = 4 - 18 + 14 = 0

The equation holds true for v = 2.

Next, let's check v = 7:

(7)² - 9(7) + 14 = 49 - 63 + 14 = 0

Again, the equation is satisfied. This confirms that both v = 2 and v = 7 are indeed the correct solutions to the quadratic equation.

Solving v² - 9v + 14 = 0 Using the Quadratic Formula

While the factoring method worked efficiently for this particular equation, the quadratic formula offers a more general approach that can be applied to any quadratic equation, regardless of its factorability. The quadratic formula is given by:

v = (-b ± √(b² - 4ac)) / 2a

where 'a', 'b', and 'c' are the coefficients of the quadratic equation in the standard form ax² + bx + c = 0. In our case, a = 1, b = -9, and c = 14. Plugging these values into the quadratic formula, we get:

v = (-(-9) ± √((-9)² - 4 * 1 * 14)) / (2 * 1) v = (9 ± √(81 - 56)) / 2 v = (9 ± √25) / 2 v = (9 ± 5) / 2

This gives us two possible solutions:

v₁ = (9 + 5) / 2 = 14 / 2 = 7 v₂ = (9 - 5) / 2 = 4 / 2 = 2

As expected, the quadratic formula yields the same solutions as the factoring method: v = 2 and v = 7.

The Discriminant: Unveiling the Nature of the Roots

Within the quadratic formula lies a crucial component known as the discriminant. The discriminant, denoted as Δ, is the expression under the square root:

Δ = b² - 4ac

The discriminant provides valuable information about the nature of the roots of a quadratic equation. By examining the value of Δ, we can determine whether the equation has:

• Two distinct real roots (Δ > 0)
• One real root (a repeated root) (Δ = 0)
• No real roots (two complex roots) (Δ < 0)

For our equation, v² - 9v + 14 = 0, the discriminant is:

Δ = (-9)² - 4 * 1 * 14 = 81 - 56 = 25

Since Δ > 0, we confirm that the equation has two distinct real roots, which we have already found to be v = 2 and v = 7.

Completing the Square: An Alternative Approach

While we have successfully solved the equation using factoring and the quadratic formula, it's worth exploring another method: completing the square. This technique involves transforming the quadratic equation into a perfect square trinomial, making it easier to isolate the variable. Let's revisit the equation v² - 9v + 14 = 0 and apply the method of completing the square.

1. Move the Constant Term: Begin by moving the constant term to the right side of the equation:

v² - 9v = -14

2. Complete the Square: To complete the square on the left side, take half of the coefficient of the linear term (-9), square it, and add it to both sides of the equation. Half of -9 is -9/2, and squaring it gives us 81/4:

v² - 9v + 81/4 = -14 + 81/4

3. Factor the Perfect Square Trinomial: The left side now forms a perfect square trinomial, which can be factored as:

(v - 9/2)² = -14 + 81/4

4. Simplify the Right Side: Combine the terms on the right side:

(v - 9/2)² = 25/4

5. Take the Square Root: Take the square root of both sides:

v - 9/2 = ±√(25/4) v - 9/2 = ±5/2

6. Isolate the Variable: Solve for v:

v = 9/2 ± 5/2

This gives us two solutions:

v₁ = (9/2) + (5/2) = 14/2 = 7 v₂ = (9/2) - (5/2) = 4/2 = 2

As with the previous methods, completing the square confirms that the solutions to the equation v² - 9v + 14 = 0 are v = 2 and v = 7.

Real-World Applications of Quadratic Equations

Quadratic equations are not merely abstract mathematical concepts; they have practical applications in various real-world scenarios. Let's explore a few examples:

• Projectile Motion: Quadratic equations are used to model the trajectory of projectiles, such as balls thrown into the air or rockets launched into space. The equation can predict the height and range of the projectile, considering factors like initial velocity and launch angle.
• Engineering: Engineers use quadratic equations in structural design to calculate the stresses and strains on beams and other structural components. This ensures the safety and stability of buildings, bridges, and other infrastructure.
• Optimization Problems: Many optimization problems, such as finding the maximum profit or minimizing costs, can be modeled using quadratic equations. The solutions to these equations provide the optimal values for the variables involved.
• Economics: Quadratic equations are used in economics to model supply and demand curves, cost functions, and profit maximization problems. This helps economists understand and predict market behavior.

Conclusion

In this comprehensive guide, we have explored the methods for solving quadratic equations, focusing on the specific example v² - 9v + 14 = 0. We have demonstrated how to solve the equation using factoring, the quadratic formula, and completing the square, each method yielding the same solutions: v = 2 and v = 7. We have also discussed the significance of the discriminant in determining the nature of the roots and highlighted the real-world applications of quadratic equations.

Mastering the techniques for solving quadratic equations is a valuable skill that will serve you well in various mathematical and scientific endeavors. By understanding the underlying principles and practicing the different methods, you will be well-equipped to tackle any quadratic equation that comes your way.

Understanding the Problem: Solving Quadratic Equations

In this article, we will address the question of how to solve for v in the quadratic equation v² - 9v + 14 = 0. This type of equation falls under the category of quadratic equations, which are polynomial equations of the second degree. They have a general form of ax² + bx + c = 0, where a, b, and c are constants, and x is the variable we aim to solve. In our specific case, a = 1, b = -9, and c = 14. Solving quadratic equations is a fundamental skill in algebra, and several methods can be employed to find the solutions, also known as roots or zeros of the equation. These solutions represent the values of v that make the equation true. This article will guide you through the factoring method, which is often the most efficient approach when applicable. We will also touch upon the quadratic formula, a universal method that can solve any quadratic equation, regardless of its factorability. By understanding these techniques, you’ll be well-equipped to tackle similar problems in the future.

Factoring the Quadratic Equation: A Step-by-Step Approach

The most straightforward method to solve the quadratic equation v² - 9v + 14 = 0 is by factoring. Factoring involves breaking down the quadratic expression into two binomial factors. To achieve this, we need to find two numbers that, when multiplied together, equal the constant term (14), and when added together, equal the coefficient of the linear term (-9). Let's call these two numbers m and n. We are looking for m and n such that:

m * n = 14

m + n = -9

By considering the factors of 14, which are 1 and 14, or 2 and 7, and keeping in mind that the sum needs to be -9, we can identify the numbers as -2 and -7. Indeed, (-2) * (-7) = 14 and (-2) + (-7) = -9. This means we can rewrite the quadratic equation as a product of two binomials:

(v - 2)(v - 7) = 0

Now, we apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, either (v - 2) = 0 or (v - 7) = 0. Solving each of these linear equations gives us the solutions for v. Factoring is a powerful technique that not only helps solve equations but also provides insights into the structure and behavior of quadratic expressions. The ability to quickly and accurately factor quadratic equations is a valuable asset in various mathematical and scientific contexts.

Solving for v: Applying the Zero-Product Property

Having factored the quadratic equation into (v - 2)(v - 7) = 0, we can now utilize the zero-product property to find the solutions for v. The zero-product property states that if the product of two or more factors is zero, then at least one of the factors must be equal to zero. In our case, this means either the factor (v - 2) is zero or the factor (v - 7) is zero. Let's consider each case separately:

1. v - 2 = 0

To solve this equation for v, we simply add 2 to both sides of the equation:

v - 2 + 2 = 0 + 2

v = 2

2. v - 7 = 0

Similarly, to solve this equation for v, we add 7 to both sides:

v - 7 + 7 = 0 + 7

v = 7

Thus, we have found two solutions for the quadratic equation v² - 9v + 14 = 0: v = 2 and v = 7. These are the values of v that, when substituted back into the original equation, will make the equation true. The zero-product property is a cornerstone in solving factored equations, providing a direct pathway from factors to solutions. It is a simple yet powerful concept that is widely used in algebra and beyond. Mastering this property is crucial for anyone working with polynomial equations and algebraic expressions.

Verification: Ensuring the Accuracy of Our Solutions

To ensure the accuracy of our solutions, it's crucial to verify them by substituting them back into the original quadratic equation, v² - 9v + 14 = 0. This step confirms that the values we found for v indeed satisfy the equation. Let's start by verifying the solution v = 2:

Substituting v = 2 into the equation, we get:

(2)² - 9(2) + 14 = 0

4 - 18 + 14 = 0

0 = 0

The equation holds true, so v = 2 is indeed a solution. Now, let's verify the solution v = 7:

Substituting v = 7 into the equation, we get:

(7)² - 9(7) + 14 = 0

49 - 63 + 14 = 0

0 = 0

Again, the equation holds true, confirming that v = 7 is also a solution. Verification is an essential practice in mathematics and problem-solving. It adds a layer of confidence to our solutions and helps catch any potential errors. By substituting our solutions back into the original equation, we confirm that they indeed satisfy the equation's conditions, ensuring the validity of our work. This practice not only reinforces our understanding of the problem but also develops a habit of critical thinking and accuracy in our mathematical endeavors.

The Quadratic Formula: A Universal Solution

While factoring is an efficient method for solving quadratic equations when applicable, it is not always possible to easily factor a given equation. In such cases, the quadratic formula provides a universal solution. The quadratic formula is a formula that provides the solutions to any quadratic equation in the form ax² + bx + c = 0, where a, b, and c are constants. The formula is given by:

v = (-b ± √(b² - 4ac)) / (2a)

In our equation, v² - 9v + 14 = 0, we have a = 1, b = -9, and c = 14. Substituting these values into the quadratic formula, we get:

v = (-(-9) ± √((-9)² - 4(1)(14))) / (2(1))

v = (9 ± √(81 - 56)) / 2

v = (9 ± √25) / 2

v = (9 ± 5) / 2

This gives us two possible solutions:

v₁ = (9 + 5) / 2 = 14 / 2 = 7

v₂ = (9 - 5) / 2 = 4 / 2 = 2

As we found earlier through factoring, the solutions to the equation are v = 2 and v = 7. The quadratic formula is a powerful tool because it can solve any quadratic equation, regardless of whether it is factorable or not. It is a fundamental concept in algebra and a must-know for anyone dealing with quadratic equations. The ability to apply the quadratic formula effectively ensures that you can always find the solutions to a quadratic equation, making it an invaluable asset in mathematical problem-solving.

The Discriminant: Understanding the Nature of Roots

Within the quadratic formula, there's a crucial component called the discriminant, which provides valuable information about the nature of the roots of a quadratic equation. The discriminant is the expression under the square root sign, which is b² - 4ac. The value of the discriminant can tell us whether the equation has two distinct real roots, one real root (a repeated root), or no real roots (two complex roots). Let's analyze the discriminant in our equation, v² - 9v + 14 = 0:

The discriminant, Δ, is calculated as:

Δ = b² - 4ac

In our case, a = 1, b = -9, and c = 14, so:

Δ = (-9)² - 4(1)(14)

Δ = 81 - 56

Δ = 25

Since the discriminant Δ is positive (25 > 0), this indicates that the quadratic equation has two distinct real roots. This aligns with our findings earlier when we solved the equation by factoring and using the quadratic formula, where we found the solutions v = 2 and v = 7. The discriminant is a powerful indicator that helps us understand the characteristics of a quadratic equation's solutions without actually solving the equation. It provides a quick way to determine the number and type of roots, which is essential in various mathematical and scientific applications. Understanding the discriminant enhances our problem-solving skills and deepens our comprehension of quadratic equations.

Conclusion: Mastering Quadratic Equation Solving

In this comprehensive guide, we've thoroughly explored the process of solving the quadratic equation v² - 9v + 14 = 0. We've demonstrated multiple methods, including factoring and the quadratic formula, and verified our solutions to ensure accuracy. Through factoring, we broke down the quadratic expression into (v - 2)(v - 7) = 0, leading us to the solutions v = 2 and v = 7. The quadratic formula provided an alternative approach, confirming the same solutions. Additionally, we discussed the discriminant, which revealed that the equation has two distinct real roots, aligning with our calculated solutions. This exploration highlights the importance of understanding various techniques for solving quadratic equations, as different methods may be more efficient depending on the specific equation. Factoring is often the quickest method when applicable, while the quadratic formula provides a universal solution. Mastering these techniques equips you with the skills to tackle a wide range of quadratic equation problems. Solving quadratic equations is a fundamental skill in algebra and has numerous applications in various fields, including physics, engineering, and economics. By understanding the underlying concepts and practicing different methods, you can confidently solve quadratic equations and apply them to real-world scenarios.