Solving Quadratic Systems: A Step-by-Step Guide
Hey everyone! Today, we're diving into the exciting world of solving systems of equations, specifically focusing on a quadratic and a linear equation. Our mission is to find the points where these equations intersect, giving us the solutions that satisfy both. We'll tackle this algebraically, ensuring we understand each step along the way. So, let's jump right into the problem: we need to determine the solution(s) for the system of equations $y=x^2-4x-21$ and $y=-5x-22$. But before we do, let's discuss what exactly systems of equations are and why they are so important in math.
Understanding Systems of Equations
So, what exactly is a system of equations? A system of equations is basically a set of two or more equations that share the same variables. Think of it as a puzzle where you need to find the values of the variables that make all the equations true at the same time. In our case, we have two equations, one quadratic and one linear, and we're looking for the x and y values that satisfy both.
Why are systems of equations so important? Well, they pop up all over the place in real-world applications. From calculating the break-even point in business to modeling the trajectory of a projectile in physics, systems of equations help us understand and solve a huge range of problems. They're a fundamental tool in mathematics and science, and mastering them opens doors to more advanced concepts.
The solutions to a system of equations are the points where the graphs of the equations intersect. This is a crucial concept, especially when we're dealing with different types of equations like our quadratic and linear system. The quadratic equation, $y = x^2 - 4x - 21$, represents a parabola, which is a U-shaped curve. The linear equation, $y = -5x - 22$, represents a straight line. The points where the parabola and the line intersect are the solutions to the system because these points satisfy both equations simultaneously.
Now, when it comes to solving systems of equations, we have a few tricks up our sleeves. Graphing is a great visual method, but it's not always precise. Substitution and elimination are algebraic methods that allow us to find exact solutions. We'll be using the substitution method today because it's particularly handy when one of the equations is already solved for one variable, like our equations are (both solved for y). So, with the basics covered, let's get back to our specific problem and solve it step-by-step.
Setting Up the Equations
Alright, let's dive into solving this system of equations. We have:
- Equation 1: $y = x^2 - 4x - 21$
- Equation 2: $y = -5x - 22$
Our goal here is to find the values of x and y that make both of these equations true. Since both equations are already solved for y, the substitution method is going to be our best friend. This method involves setting the expressions for y from both equations equal to each other. Why? Because if both equations are equal to y, then they must also be equal to each other. Makes sense, right?
So, let's go ahead and do that. We'll set the right-hand side of Equation 1 equal to the right-hand side of Equation 2: $x^2 - 4x - 21 = -5x - 22$. This step is crucial because it combines our two equations into a single equation with just one variable, x. Now, we have a quadratic equation to solve, which is something we can definitely handle.
The next step is to rearrange this equation into the standard quadratic form, which is $ax^2 + bx + c = 0$. This form is super important because it allows us to use various methods for solving quadratic equations, such as factoring, completing the square, or the quadratic formula. Getting our equation into this standard form is like setting the stage for the grand finale – finding the solutions!
To get there, we need to move all the terms to one side of the equation, leaving zero on the other side. We can do this by adding $5x$ to both sides and adding $22$ to both sides. This ensures we maintain the balance of the equation, keeping the left side equal to the right side. It's like a mathematical balancing act! So, let's perform these operations and see what our new equation looks like. Once we have it in the standard form, we'll be ready to roll up our sleeves and solve for x.
Solving the Quadratic Equation
Okay, after moving all the terms to one side, we get: $x^2 - 4x - 21 + 5x + 22 = 0$.
Now, let's simplify this by combining like terms. We have a $–4x$ and a $+5x$, which combine to give us $+x$. And we have a $-21$ and a $+22$, which combine to give us $+1$. So, our equation now looks like this: $x^2 + x + 1 = 0$. Ta-da! We've successfully transformed our equation into the standard quadratic form: $ax^2 + bx + c = 0$, where $a = 1$, $b = 1$, and $c = 1$.
Now comes the exciting part: solving for x. We have a few options here, but since this quadratic equation doesn't seem to factor easily, we're going to use the quadratic formula. The quadratic formula is a powerful tool that can solve any quadratic equation, no matter how messy it looks. It's like a universal key that unlocks the solutions to quadratic puzzles. The formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. It might look a bit intimidating at first, but trust me, it's not as scary as it seems.
To use the quadratic formula, we just need to plug in the values of $a$, $b$, and $c$ from our equation. In our case, $a = 1$, $b = 1$, and $c = 1$. So, let's substitute these values into the formula and see what we get: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}$.
Now, we need to simplify this expression. First, let's simplify the expression under the square root. We have $1^2$, which is just 1, and then we have $-4(1)(1)$, which is $-4$. So, under the square root, we have $1 - 4$, which equals $-3$. This is where things get interesting. We have a negative number under the square root. What does that mean for our solutions? Let's explore that in the next section.
Analyzing the Discriminant
Alright, we've reached a crucial point in our solution. We've plugged the values into the quadratic formula and simplified it to $x = \frac{-1 \pm \sqrt{-3}}{2}$. Notice that we have a square root of a negative number, specifically $\sqrt{-3}$. This is a big deal in the world of real numbers because you can't take the square root of a negative number and get a real number result. Think about it: any real number squared is always positive or zero. So, there's no real number that, when multiplied by itself, gives you $-3$.
This situation tells us something very important about the solutions to our quadratic equation, and consequently, about the solutions to our system of equations. The part of the quadratic formula under the square root, $b^2 - 4ac$, is called the discriminant. The discriminant is like a detective that gives us clues about the nature of the solutions. If the discriminant is positive, we have two distinct real solutions. If it's zero, we have one real solution (a repeated root). And, as we've seen, if it's negative, we have no real solutions.
In our case, the discriminant is $-3$, which is negative. This means that our quadratic equation has no real solutions. Graphically, this means that the parabola and the line in our system of equations do not intersect. They might come close, but they never actually cross paths. This is a key insight because it directly answers our original question. If the quadratic equation has no real solutions, then the system of equations also has no real solutions.
So, what does this mean for the answer choices? We can confidently eliminate any options that list specific solution points because we know there are no real solutions. The only option that remains is the one that states there are no solutions. It's like solving a mystery and finding out the treasure chest is empty – a bit disappointing, but still a valid solution!
Final Answer and Implications
Therefore, the system of equations $y = x^2 - 4x - 21$ and $y = -5x - 22$ has no solutions. This corresponds to option D in our multiple-choice options. We arrived at this conclusion by using the substitution method to combine the equations, applying the quadratic formula to solve for x, and then analyzing the discriminant to determine the nature of the solutions.
This result highlights an important concept in solving systems of equations: not all systems have solutions. Sometimes, the equations represent lines or curves that never intersect, leading to no solution. Understanding how to identify these cases is just as crucial as knowing how to find solutions when they exist. The discriminant is a powerful tool in this regard, allowing us to quickly determine whether a quadratic equation has real solutions without going through the entire solving process.
Moreover, this exercise demonstrates the interconnectedness of different mathematical concepts. We used the substitution method, the quadratic formula, and the concept of the discriminant, all in one problem. This is typical in mathematics, where different ideas often come together to solve complex problems. It's like having a toolbox full of tools, and knowing which tool to use for each specific task.
So, next time you encounter a system of equations, remember the steps we've taken today: substitute, simplify, apply the quadratic formula (if needed), and analyze the discriminant. And remember, it's okay if there are no solutions – that's a solution in itself!
Practice Problems
To solidify your understanding of solving systems of equations, especially those involving quadratic equations, let's tackle a few practice problems. These exercises will help you apply the techniques we've discussed and build your confidence in handling different scenarios. Remember, practice makes perfect!
Practice Problem 1:
Solve the following system of equations:
Practice Problem 2:
Determine the solution(s) of the system:
Practice Problem 3:
Find the intersection points of the equations:
For each problem, follow the steps we've outlined: set the equations equal to each other, simplify to get a quadratic equation in standard form, use the quadratic formula if necessary, and analyze the discriminant to determine the nature of the solutions. Don't be afraid to make mistakes – they're a natural part of the learning process. The key is to learn from them and keep practicing.
Solving systems of equations is a fundamental skill in algebra, and mastering it will open doors to more advanced topics in mathematics and other fields. So, grab a pencil and paper, and let's get practicing! Happy solving!
