Solving Quadratic Equations X^2 + Bx + C = 0 With A Unique Solution

Introduction

In the realm of mathematics, quadratic equations hold a pivotal position. These equations, characterized by the general form ax^2 + bx + c = 0, where a, b, and c are constants, and x is the variable, appear frequently in various mathematical and real-world problems. Understanding how to solve quadratic equations is a fundamental skill in algebra. One common method for solving quadratic equations involves finding the roots, or solutions, of the equation. These roots represent the values of x that satisfy the equation. In some cases, a quadratic equation may have two distinct real roots, while in other cases, it may have one repeated real root or two complex roots. When a quadratic equation has only one solution, it indicates that the discriminant (b^2 - 4ac) is equal to zero, and the quadratic equation represents a perfect square trinomial.

This article delves into a specific scenario where the quadratic equation x^2 + bx + c = 0 has only one solution, x = -15. Our primary objective is to determine the values of the coefficients b and c. This exploration will not only reinforce the understanding of quadratic equations but also showcase how specific conditions can lead to unique solutions. By employing algebraic techniques and leveraging the properties of quadratic equations, we aim to unravel the values of b and c that satisfy the given condition. This problem serves as a practical example of how mathematical principles can be applied to solve concrete problems, highlighting the elegance and utility of quadratic equations in problem-solving.

Problem Statement

We are given a quadratic equation in the form x^2 + bx + c = 0, where b and c are unknown constants. The crucial piece of information provided is that this equation has only one solution, which is x = -15. Our task is to find the values of b and c that satisfy this condition. This problem is intriguing because it presents a constraint – the uniqueness of the solution – which we must use to our advantage. The fact that there is only one solution implies that the quadratic equation is a perfect square trinomial. This is because a perfect square trinomial, when factored, results in the square of a binomial, leading to a single repeated root. Understanding this implication is key to solving the problem efficiently. By recognizing this, we can employ techniques such as completing the square or using the relationship between the roots and coefficients of a quadratic equation to determine the values of b and c. This problem not only tests our algebraic skills but also our ability to recognize patterns and apply appropriate strategies in problem-solving. The solution will demonstrate how a combination of algebraic manipulation and conceptual understanding can lead to a concise and elegant answer.

Method 1: Factoring and Perfect Square Trinomials

One effective method to tackle this problem involves recognizing that a quadratic equation with a single solution must be a perfect square trinomial. A perfect square trinomial is a trinomial that can be factored into the square of a binomial. In the context of quadratic equations, this means that the equation can be expressed in the form (x + k)^2 = 0, where k is a constant. When expanded, this form yields x^2 + 2kx + k^2 = 0. This form is particularly useful because it directly relates the coefficients of the quadratic equation to the single root. Given that the only solution to our equation x^2 + bx + c = 0 is x = -15, we can deduce that the quadratic equation must be of the form (x + 15)^2 = 0. This is because substituting x = -15 into (x + 15)^2 results in 0, satisfying the equation. Now, our task is to expand (x + 15)^2 and compare the resulting coefficients with the original equation x^2 + bx + c = 0. Expanding (x + 15)^2 gives us x^2 + 30x + 225. By comparing the coefficients of this expanded form with the original equation, we can directly determine the values of b and c. The coefficient of x in the expanded form is 30, which corresponds to b in the original equation. Similarly, the constant term in the expanded form is 225, which corresponds to c in the original equation. Therefore, by leveraging the concept of perfect square trinomials and expanding the appropriate form, we can efficiently find the values of b and c that satisfy the given condition. This method highlights the importance of recognizing patterns and utilizing specific algebraic forms to simplify problem-solving.

Method 2: Using the Discriminant

The discriminant, denoted as Δ, is a crucial component of the quadratic formula and provides valuable information about the nature of the roots of a quadratic equation. For a quadratic equation in the form ax^2 + bx + c = 0, the discriminant is defined as Δ = b^2 - 4ac. The value of the discriminant determines whether the quadratic equation has two distinct real roots, one repeated real root, or two complex roots. Specifically, if Δ > 0, the equation has two distinct real roots; if Δ = 0, the equation has one repeated real root; and if Δ < 0, the equation has two complex roots. In our problem, we are given that the quadratic equation x^2 + bx + c = 0 has only one solution, which implies that the discriminant must be equal to zero. This is because a single solution indicates a repeated root, which occurs when the discriminant is zero. Applying this condition to our equation, we have Δ = b^2 - 4(1)(c) = 0, which simplifies to b^2 - 4c = 0. This equation provides a relationship between b and c, but we need another equation to solve for the individual values of b and c. Since we know that the single solution is x = -15, we can substitute this value into the original equation x^2 + bx + c = 0. This gives us (-15)^2 + b(-15) + c = 0, which simplifies to 225 - 15b + c = 0. Now we have a system of two equations: 1) b^2 - 4c = 0 and 2) 225 - 15b + c = 0. We can solve this system of equations to find the values of b and c. One way to do this is to solve the second equation for c, which gives us c = 15b - 225, and then substitute this expression for c into the first equation. This will result in a quadratic equation in terms of b only, which we can solve using factoring or the quadratic formula. Once we find the value(s) of b, we can substitute them back into the equation c = 15b - 225 to find the corresponding value(s) of c. This method demonstrates how the discriminant can be used to understand the nature of the roots and how the condition of a single solution can be leveraged to set up a system of equations, ultimately leading to the determination of the unknown coefficients.

Step-by-Step Solution Using Method 2

To provide a clearer understanding of how to apply Method 2, let's walk through a step-by-step solution. This will illustrate the process of using the discriminant and solving the resulting system of equations. First, recall that the discriminant for a quadratic equation ax^2 + bx + c = 0 is given by Δ = b^2 - 4ac. In our case, the equation is x^2 + bx + c = 0, so a = 1. Since the equation has only one solution, the discriminant must be zero, meaning b^2 - 4(1)(c) = 0, which simplifies to b^2 - 4c = 0 (Equation 1). Next, we use the information that x = -15 is the only solution. Substituting this value into the original equation gives us (-15)^2 + b(-15) + c = 0, which simplifies to 225 - 15b + c = 0 (Equation 2). Now we have a system of two equations with two unknowns, b and c. We can solve this system by first solving Equation 2 for c: c = 15b - 225. Then, we substitute this expression for c into Equation 1: b^2 - 4(15b - 225) = 0. This simplifies to b^2 - 60b + 900 = 0. This is a quadratic equation in terms of b, which we can solve by factoring. Notice that the equation is a perfect square trinomial: (b - 30)^2 = 0. Therefore, the only solution for b is b = 30. Now that we have the value of b, we can substitute it back into the equation c = 15b - 225 to find c: c = 15(30) - 225 = 450 - 225 = 225. Thus, the values of b and c that satisfy the given conditions are b = 30 and c = 225. This step-by-step solution demonstrates how the discriminant, combined with the given solution, allows us to set up and solve a system of equations, ultimately leading to the determination of the unknown coefficients. The clarity of each step ensures that the method is easily understood and can be applied to similar problems.

Method 3: Vieta's Formulas

Vieta's formulas provide a powerful connection between the roots of a polynomial equation and its coefficients. For a quadratic equation in the form ax^2 + bx + c = 0, Vieta's formulas state that the sum of the roots is equal to -b/a and the product of the roots is equal to c/a. In our case, the quadratic equation is x^2 + bx + c = 0, so a = 1. We are given that the equation has only one solution, x = -15. This means that both roots of the quadratic equation are the same, and they are both equal to -15. Let's denote the roots as x_1 and x_2. According to Vieta's formulas, the sum of the roots is x_1 + x_2 = -b/a, and the product of the roots is x_1 * x_2 = c/a. Since x_1 = x_2 = -15 and a = 1, we can substitute these values into Vieta's formulas. The sum of the roots is (-15) + (-15) = -30, so we have -30 = -b/1, which implies b = 30. The product of the roots is (-15) * (-15) = 225, so we have 225 = c/1, which implies c = 225. Therefore, using Vieta's formulas, we can directly determine the values of b and c without solving a system of equations. This method highlights the elegance and efficiency of Vieta's formulas in relating the roots and coefficients of a polynomial equation. By understanding and applying these formulas, we can often simplify the process of solving problems involving quadratic equations and their roots. The directness of this method makes it a valuable tool in our problem-solving arsenal.

Comparing the Methods

In this exploration of the quadratic equation x^2 + bx + c = 0 with a single solution x = -15, we employed three distinct methods to determine the values of b and c. Each method offers a unique perspective and approach, highlighting the multifaceted nature of mathematical problem-solving. Method 1, which involved factoring and recognizing perfect square trinomials, leverages the fundamental concept that a quadratic equation with a single solution can be expressed as the square of a binomial. This method is intuitive and straightforward, requiring the expansion of (x + 15)^2 and comparison of coefficients. It emphasizes the importance of pattern recognition and algebraic manipulation. Method 2 utilized the discriminant, a key concept in quadratic equations that determines the nature of the roots. By setting the discriminant b^2 - 4ac to zero, we established a relationship between b and c. Combining this with the equation obtained by substituting x = -15 into the original quadratic equation, we formed a system of two equations. Solving this system yielded the values of b and c. This method showcases the power of the discriminant in analyzing quadratic equations and the utility of solving systems of equations. Method 3, employing Vieta's formulas, provided the most direct and elegant solution. Vieta's formulas establish a direct link between the roots of a polynomial equation and its coefficients. By applying these formulas, we were able to determine the values of b and c almost immediately. This method underscores the efficiency of using established mathematical relationships to solve problems. Comparing these methods, we find that while all three lead to the correct answer (b = 30 and c = 225), they differ in their approach and complexity. Method 1 is conceptually simple but requires algebraic manipulation. Method 2 is more analytical, involving the discriminant and a system of equations. Method 3 is the most concise, leveraging Vieta's formulas for a direct solution. The choice of method may depend on personal preference, familiarity with the concepts, and the specific context of the problem. Ultimately, understanding multiple methods enhances problem-solving skills and provides a deeper appreciation for the interconnectedness of mathematical concepts.

Conclusion

In conclusion, we have successfully determined the values of b and c for the quadratic equation x^2 + bx + c = 0, given that it has only one solution, x = -15. Through the application of three distinct methods – factoring perfect square trinomials, using the discriminant, and employing Vieta's formulas – we arrived at the consistent result: b = 30 and c = 225. This problem not only reinforces our understanding of quadratic equations but also highlights the versatility of algebraic techniques in problem-solving. The fact that we could approach the problem from multiple angles, each with its own unique perspective, underscores the richness and interconnectedness of mathematical concepts. Each method offered valuable insights. Factoring perfect square trinomials emphasized the relationship between the single solution and the structure of the quadratic equation. The discriminant method showcased its utility in determining the nature of the roots and setting up a system of equations. Vieta's formulas provided a direct and elegant solution, demonstrating the power of established mathematical relationships. The consistency of the results across these methods further validates our solution and reinforces our confidence in the applied techniques. Moreover, this exploration serves as a reminder that effective problem-solving often involves choosing the most appropriate method based on the given information and the desired level of efficiency. The journey through these methods has not only provided a solution to the specific problem but also enhanced our mathematical toolkit and problem-solving acumen. The values of b = 30 and c = 225 complete the quadratic equation, which can be expressed as x^2 + 30x + 225 = 0, or equivalently, (x + 15)^2 = 0. This final form clearly demonstrates why x = -15 is the only solution, solidifying our understanding of the problem and its solution.