Finding Zeros: Solving Quadratic Equations Explained

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Hey everyone! Today, we're diving deep into the world of quadratic functions, specifically how to determine the zeros of a quadratic function like y=βˆ’x2+16xβˆ’39y = -x^2 + 16x - 39. Finding the zeros is essentially finding the x-values where the function crosses the x-axis, or where y=0y = 0. This is a super important skill in algebra, and it's something that will pop up again and again. So, let's break it down in a way that's easy to understand. We'll explore different methods to find these zeros, ensuring you have a solid grasp of the concepts and can tackle similar problems with confidence. Getting comfortable with these techniques is a game-changer for acing your math tests and understanding how these functions work in real-world scenarios. It's all about breaking down the problem into smaller, manageable steps. By the end, you'll be able to not only find the zeros but also understand the why behind each step.

So, what exactly are these "zeros" we keep talking about? In simpler terms, the zeros of a quadratic function are the x-values that make the function equal to zero. These are also known as the roots or x-intercepts of the equation. Graphically, they represent the points where the parabola (the U-shaped curve of the quadratic function) intersects the x-axis. Each zero gives us a direct insight into the behavior of the quadratic function. The real zeros are where the graph of the function crosses the x-axis. If a quadratic equation has no real zeros, then the parabola does not cross the x-axis. Instead, it either sits entirely above or below the x-axis. Identifying these points is key to analyzing the function's behavior. We can understand the function's behavior by knowing where these zeros are located. Let's get started, and I'll walk you through this process step by step!

Understanding Quadratic Functions and Zeros

Alright, before we get our hands dirty with the equation, let's quickly recap what a quadratic function actually is. A quadratic function is a function that can be written in the form y=ax2+bx+cy = ax^2 + bx + c, where a, b, and c are constants, and ae0a e 0. The graph of a quadratic function is a parabola. The zeros of the quadratic function are the solutions to the equation ax2+bx+c=0ax^2 + bx + c = 0. In other words, they are the x-values for which y=0y=0. These zeros are also often called roots or x-intercepts, since they are the points where the graph intersects the x-axis. Remember that a quadratic function can have zero, one, or two real zeros, depending on the discriminant. The zeros tell us a lot about the function's graph; they tell us where the function crosses the x-axis. When we talk about "solving" a quadratic equation, we're really just trying to find these zeros. These are the values of x that satisfy the equation. There are multiple methods to find these values, but each method serves the same purpose: to identify the points where the quadratic function crosses the x-axis. Knowing this is important for various reasons, including modeling real-world problems and understanding how these equations behave in the context of a graph.

Now, let's talk about the specific quadratic equation we're working with: y=βˆ’x2+16xβˆ’39y = -x^2 + 16x - 39. The goal here is to find the values of x that make y equal to zero. This means we're trying to solve the equation βˆ’x2+16xβˆ’39=0-x^2 + 16x - 39 = 0. The coefficients in this equation are: a=βˆ’1a = -1, b=16b = 16, and c=βˆ’39c = -39. These values will be useful in the methods we use to solve for the zeros, especially the quadratic formula. Understanding these basics sets the stage for solving more complex problems later on. Each method offers a unique way to approach finding the zeros, and understanding the logic behind them strengthens your overall grasp of quadratic functions. Let's now explore the various methods we can use to determine the zeros of the quadratic function.

Methods for Finding Zeros

There are several methods you can use to determine the zeros of a quadratic function. Let's walk through three common approaches: factoring, completing the square, and using the quadratic formula. Each method has its pros and cons, and which one you choose might depend on the specific form of the equation or your personal preference. So, let’s get started, shall we?

1. Factoring

Factoring is often the easiest method if the quadratic expression can be factored easily. The goal is to rewrite the quadratic equation as a product of two binomials. If we can factor the quadratic equation into the form (xβˆ’p)(xβˆ’q)=0(x - p)(x - q) = 0, then the zeros are simply x=px = p and x=qx = q. For our equation, βˆ’x2+16xβˆ’39=0-x^2 + 16x - 39 = 0, it might be helpful to multiply both sides by -1 to make the leading coefficient positive, giving us x2βˆ’16x+39=0x^2 - 16x + 39 = 0. Now, we need to find two numbers that multiply to 39 and add up to -16. These numbers are -3 and -13. So, we can factor the equation as (xβˆ’3)(xβˆ’13)=0(x - 3)(x - 13) = 0. Setting each factor equal to zero gives us: xβˆ’3=0x - 3 = 0 which gives us x=3x = 3 and xβˆ’13=0x - 13 = 0 which gives us x=13x = 13. Thus, the zeros of the function are 3 and 13. Remember that this method only works if the quadratic expression is easily factorable. When factoring, you must ensure that the product of the two binomials correctly expands to match the original quadratic equation. Factoring is a handy way to solve for zeros because it directly reveals the x-intercepts, making it easier to visualize the graph of the function.

2. Completing the Square

Completing the square is a reliable method that works for any quadratic equation, although it can be a bit more involved. The method involves manipulating the equation to create a perfect square trinomial on one side. First, start with the original equation, βˆ’x2+16xβˆ’39=0-x^2 + 16x - 39 = 0. It's easier to work with a positive leading coefficient, so we multiply by -1 to get x2βˆ’16x+39=0x^2 - 16x + 39 = 0. Now, isolate the x terms: x2βˆ’16x=βˆ’39x^2 - 16x = -39. To complete the square, take half of the coefficient of the x term (-16), square it ((βˆ’16/2)2=64)((-16/2)^2 = 64), and add it to both sides: x2βˆ’16x+64=βˆ’39+64x^2 - 16x + 64 = -39 + 64. This simplifies to (xβˆ’8)2=25(x - 8)^2 = 25. Now, take the square root of both sides: xβˆ’8=Β±5x - 8 = \pm 5. Finally, solve for x: x=8Β±5x = 8 \pm 5. This gives us two solutions: x=8+5=13x = 8 + 5 = 13 and x=8βˆ’5=3x = 8 - 5 = 3. Hence, we find the zeros to be 3 and 13. Completing the square is perfect when factoring feels tricky. It always works and also helps reveal the vertex of the parabola. Mastering completing the square equips you with a powerful tool for solving all types of quadratic equations.

3. Quadratic Formula

The quadratic formula is the ultimate go-to solution for finding the zeros of a quadratic equation. It works for all quadratic equations, regardless of whether they can be factored easily. The quadratic formula is given by: x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. For our equation, βˆ’x2+16xβˆ’39=0-x^2 + 16x - 39 = 0, we have a=βˆ’1a = -1, b=16b = 16, and c=βˆ’39c = -39. Plugging these values into the formula, we get: x=βˆ’16Β±162βˆ’4(βˆ’1)(βˆ’39)2(βˆ’1)x = \frac{-16 \pm \sqrt{16^2 - 4(-1)(-39)}}{2(-1)}. This simplifies to x=βˆ’16Β±256βˆ’156βˆ’2x = \frac{-16 \pm \sqrt{256 - 156}}{-2}, which further simplifies to x=βˆ’16Β±100βˆ’2x = \frac{-16 \pm \sqrt{100}}{-2}. That becomes x=βˆ’16Β±10βˆ’2x = \frac{-16 \pm 10}{-2}. Now we have two solutions: x=βˆ’16+10βˆ’2=βˆ’6βˆ’2=3x = \frac{-16 + 10}{-2} = \frac{-6}{-2} = 3 and x=βˆ’16βˆ’10βˆ’2=βˆ’26βˆ’2=13x = \frac{-16 - 10}{-2} = \frac{-26}{-2} = 13. Thus, the zeros are 3 and 13. The quadratic formula guarantees a solution, and you should use it whenever in doubt. This formula provides a straightforward way to calculate the zeros of a quadratic equation and is a must-know for every algebra student. This is the most reliable method for finding the zeros, and once you master this formula, you can solve any quadratic equation effortlessly.

Conclusion: Zeros Found!

There you have it! We've successfully determined the zeros of the quadratic function y=βˆ’x2+16xβˆ’39y = -x^2 + 16x - 39 using three different methods: factoring, completing the square, and the quadratic formula. The zeros, or roots, are x = 3 and x = 13. Each method offers a unique approach, allowing you to choose the one that best suits the problem at hand. Whether you prefer the simplicity of factoring or the guaranteed solution provided by the quadratic formula, the ability to find the zeros of a quadratic function is an invaluable skill. These zeros represent the x-intercepts of the parabola, providing key insights into the function's behavior. Understanding the zeros is crucial for graphing and analyzing quadratic functions, and it builds a solid foundation for more complex mathematical concepts.

Keep practicing these methods, and soon you'll be finding the zeros of any quadratic equation with ease! Understanding the concept of zeros and the various methods for finding them is a key step in mastering algebra and other higher-level math concepts. Good luck, and keep up the great work!