Solve X² = 2x + 3 Graphically Identifying The Correct System Of Equations

by James Vasile 74 views

Hey guys! Let's dive into this math problem where we're trying to figure out which system of equations we can graph to solve the equation x² = 2x + 3. It might sound a bit complicated at first, but trust me, we'll break it down and make it super clear. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the options, let's make sure we understand what the question is really asking. We have a quadratic equation: x² = 2x + 3. Our mission, should we choose to accept it, is to find a system of equations that, when graphed, will show us the solutions to this equation. What does that even mean? Well, when we talk about solving an equation graphically, we're looking for the points where two graphs intersect. These intersection points represent the values of x and y that satisfy both equations simultaneously. So, we need to rewrite our given equation in a way that splits it into two separate equations that we can graph.

The key here is to think about how we can manipulate the original equation to get two equations in the form of y = something. This is because when we graph equations, we typically graph them in the Cartesian plane, where we have an x-axis and a y-axis. Equations in the form of y = f(x) are easy to graph because for every x value, we can find a corresponding y value. So, let's keep this in mind as we evaluate our options.

Now, let's think step-by-step about how we can transform the given equation x² = 2x + 3 into a system of two equations. The most straightforward approach is to recognize that we can treat each side of the equation as a separate function. One function can be y = x², representing a parabola, and the other can be y = 2x + 3, representing a straight line. The solutions to the original equation will be the x-coordinates of the points where these two graphs intersect. This is a crucial concept, so make sure it's crystal clear before we move on. We're essentially turning one equation into a visual problem, which can often make it easier to solve.

Another way to think about it is by rearranging the original equation to have zero on one side. This is a common technique in algebra. If we subtract 2x and 3 from both sides, we get x² - 2x - 3 = 0. Now, we can think about graphing y = x² - 2x - 3, which is a parabola, and y = 0, which is the x-axis. The solutions to the equation will be the x-intercepts of the parabola, which are the points where the parabola intersects the x-axis. This is another valid way to break down the problem and visualize the solutions. So, with these ideas in mind, let's take a look at the given options and see which one matches our strategy.

Evaluating the Options

Okay, let's roll up our sleeves and dive into the options! We've got a few different systems of equations to consider, and we need to figure out which one can be graphed to find the solutions to our original equation, x² = 2x + 3. Remember, our goal is to find a system where the intersection points of the graphs will give us the values of x that satisfy the equation. So, let's put on our detective hats and analyze each option carefully.

Option 1:

{y = x² + 2x + 3 \ y = 2x + 3}

This first system looks interesting. The first equation, y = x² + 2x + 3, is a quadratic equation, which means its graph will be a parabola. The second equation, y = 2x + 3, is a linear equation, so its graph will be a straight line. Now, the question is, does this system directly relate to our original equation, x² = 2x + 3? Let's think about it. If we set the two y's equal to each other (since at the intersection points, the y-values will be the same), we get x² + 2x + 3 = 2x + 3. If we simplify this by subtracting 2x + 3 from both sides, we end up with x² = 0. This is not the same as our original equation, x² = 2x + 3. So, while this system of equations will have solutions (where the parabola and the line intersect), they won't be the solutions to our original problem. Therefore, this option is not the correct one. We need to find a system that, when we equate the y-values, gives us back our original equation.

Option 2:

{y = x² - 3 \ y = 2x + 3}

Alright, let's move on to the second option. This system also has a quadratic equation, y = x² - 3, and a linear equation, y = 2x + 3. Again, we have a parabola and a straight line. To see if this system works, let's do the same thing we did before: set the two y's equal to each other. This gives us x² - 3 = 2x + 3. Now, let's rearrange this equation to see if it matches our original equation. If we add 3 to both sides, we get x² = 2x + 6. Hmm, this is close, but it's not quite the same as x² = 2x + 3. The constant term is different (6 instead of 3). So, this system of equations, while interesting, also doesn't lead us to the solutions of our initial equation. It's like we're on the right track, but we're just a little bit off. We need to keep searching for the perfect match.

Option 3:

{y = x² - 2x - 3 \ y = 0}

Now, let's consider the third option. This system has a quadratic equation, y = x² - 2x - 3, and a very simple equation, y = 0. This is actually a clever setup! Remember how we talked about rearranging the original equation to have zero on one side? That's exactly what's happening here. The equation y = 0 represents the x-axis. So, we're looking for the points where the parabola y = x² - 2x - 3 intersects the x-axis. To see if this works, let's set the two y's equal to each other: x² - 2x - 3 = 0. Now, let's go back to our original equation, x² = 2x + 3. If we subtract 2x and 3 from both sides, we get x² - 2x - 3 = 0. Bingo! This is exactly the equation we got from setting the y's equal in our current system. This means that the solutions to this system of equations will indeed be the solutions to our original equation. The x-intercepts of the parabola y = x² - 2x - 3 will give us the x-values that satisfy x² = 2x + 3. So, it looks like we have a winner! This is the system we can graph to find the solutions.

The Correct Answer

Alright, after carefully analyzing each option, we've arrived at the answer! The system of equations that can be graphed to find the solution(s) to x² = 2x + 3 is:

{y = x² - 2x - 3 \ y = 0}

This is because when we set the two equations equal to each other, we get x² - 2x - 3 = 0, which is a rearranged form of our original equation. The solutions to this system will be the x-intercepts of the parabola y = x² - 2x - 3, which correspond to the values of x that satisfy x² = 2x + 3. So, we've successfully navigated through the problem and found the correct system of equations. Give yourselves a pat on the back, guys! You've done some awesome math detective work today.

Key Takeaways

Before we wrap up, let's quickly recap the key concepts we've learned in this problem. Understanding these takeaways will not only help you with similar problems but also deepen your understanding of how equations and graphs are related. So, let's highlight the main points:

  • Graphical Solutions: We learned that solving an equation graphically involves finding the intersection points of the graphs of the equations in a system. The x-coordinates of these intersection points give us the solutions to the original equation. This is a powerful way to visualize algebraic solutions.
  • Rearranging Equations: We saw how important it is to rearrange equations to match the form we need. In this case, we rearranged x² = 2x + 3 to x² - 2x - 3 = 0. This allowed us to identify the correct system of equations for graphical solution.
  • Systems of Equations: We understood that a single equation can be represented as a system of two equations by treating each side of the equation as a separate function (y = f(x)). This is a fundamental technique in algebra and calculus.
  • Quadratic Equations and Parabolas: We reinforced our understanding of quadratic equations and their graphical representation as parabolas. Knowing the shape of a parabola helps us visualize the solutions and intercepts.
  • Linear Equations and Straight Lines: Similarly, we worked with linear equations and their graphical representation as straight lines. The intersection of a parabola and a line can give us the solutions to a quadratic equation.

By mastering these concepts, you'll be well-equipped to tackle a wide range of math problems, especially those involving equations and graphs. Keep practicing, keep exploring, and remember that math can be both challenging and incredibly rewarding!

Solving Equations Graphically Find the System for x² = 2x + 3