Solving For Roots Graphically Polynomial Equation Example

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Hey guys! Let's dive into how we can find the roots of a polynomial equation by graphing. We've got a cool problem here: Sandra graphed a system of equations to solve x3βˆ’2x2βˆ’11x+12=x3βˆ’13xβˆ’12x^3 - 2x^2 - 11x + 12 = x^3 - 13x - 12. Our mission is to figure out the roots of this polynomial equation. Spoiler alert: the roots are the x-values where the graph crosses the x-axis, or where the two graphed equations intersect. Buckle up, let's get started!

Understanding the Problem

Before we jump into graphing, let's make sure we really understand what we're trying to do. So, the polynomial equation we're dealing with is x3βˆ’2x2βˆ’11x+12=x3βˆ’13xβˆ’12x^3 - 2x^2 - 11x + 12 = x^3 - 13x - 12. We need to find the values of xx that make this equation true. These values are also known as the roots or zeros of the polynomial. Graphically, these roots are where the function's graph intersects the x-axis. Now, Sandra used a clever trick by graphing a system of equations. This means she probably separated the original equation into two simpler equations and graphed them separately. The points where these graphs intersect give us the solutions to the original equation. This is because at the intersection points, the y-values of both equations are equal, which satisfies our original equation. The beauty of this method lies in visualizing the solutions, making it easier to grasp compared to algebraic manipulation alone. Understanding this sets the stage for efficiently finding the roots and appreciating the graphical approach.

Transforming the Equation

Okay, let's start by simplifying our equation a bit. We have x3βˆ’2x2βˆ’11x+12=x3βˆ’13xβˆ’12x^3 - 2x^2 - 11x + 12 = x^3 - 13x - 12. To make things easier, we can rearrange the equation so that all terms are on one side, and we have zero on the other. This is a standard practice when solving polynomial equations because it sets us up to find the roots more easily. Let’s subtract x3x^3 from both sides: βˆ’2x2βˆ’11x+12=βˆ’13xβˆ’12-2x^2 - 11x + 12 = -13x - 12. Now, let's add 13x13x and 1212 to both sides to get everything on the left: βˆ’2x2βˆ’11x+13x+12+12=0-2x^2 - 11x + 13x + 12 + 12 = 0. Simplify it down, and we get: βˆ’2x2+2x+24=0-2x^2 + 2x + 24 = 0. To make it even simpler, we can divide the entire equation by -2: x2βˆ’xβˆ’12=0x^2 - x - 12 = 0. Now we have a quadratic equation that’s much easier to work with. This transformation is crucial because it converts a complex-looking equation into a manageable form, ready for graphing or factoring. By reducing the equation to its simplest form, we pave the way for a more straightforward solution process, ensuring we can accurately identify the roots.

Factoring the Quadratic Equation

Now that we have the simplified quadratic equation, x2βˆ’xβˆ’12=0x^2 - x - 12 = 0, let's factor it. Factoring is a classic method to find the roots of a quadratic equation. We're looking for two numbers that multiply to -12 and add up to -1 (the coefficient of our x term). Think about it – what two numbers fit this description? Bingo! -4 and 3 work perfectly because (-4) * 3 = -12 and (-4) + 3 = -1. So, we can rewrite our equation in factored form as: (xβˆ’4)(x+3)=0(x - 4)(x + 3) = 0. Factoring is a powerful technique because it breaks down the equation into manageable parts, each representing a potential root. This step is essential because it directly reveals the values of x that make the equation zero, which are precisely the roots we're after. By transforming the quadratic equation into its factored form, we're setting up a straightforward path to identify the solutions, ensuring accuracy and efficiency in our root-finding mission.

Finding the Roots

We're almost there, guys! Now that we've factored the quadratic equation into (xβˆ’4)(x+3)=0(x - 4)(x + 3) = 0, finding the roots is the easy part. Remember, the roots are the values of xx that make the equation true. For the product of two factors to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for xx:

  1. xβˆ’4=0x - 4 = 0 gives us x=4x = 4
  2. x+3=0x + 3 = 0 gives us x=βˆ’3x = -3

So, the roots of the equation are x=4x = 4 and x=βˆ’3x = -3. These are the points where the graph of the equation would intersect the x-axis. In the context of Sandra's problem, these values are also the x-coordinates where the graphs of her two original equations intersect. Finding the roots is the culmination of our efforts, providing the solutions that satisfy the equation. This step is critical because it directly answers the question, identifying the specific values of x that make the polynomial equation true. By setting each factor to zero, we efficiently uncover the roots, ensuring a clear and precise understanding of the equation's solutions.

Visualizing the Solution

Let's take a step back and visualize what we've done. Sandra started with x3βˆ’2x2βˆ’11x+12=x3βˆ’13xβˆ’12x^3 - 2x^2 - 11x + 12 = x^3 - 13x - 12. She then graphed a system of equations. A likely way she did this was by setting y=x3βˆ’2x2βˆ’11x+12y = x^3 - 2x^2 - 11x + 12 and y=x3βˆ’13xβˆ’12y = x^3 - 13x - 12, and then graphed these two cubic equations. The points where these two graphs intersect are the solutions to the original equation because, at those points, the y-values (and thus the expressions) are equal. We simplified the equation to x2βˆ’xβˆ’12=0x^2 - x - 12 = 0, which is a parabola when graphed. The roots of this quadratic equation, which we found to be x=4x = 4 and x=βˆ’3x = -3, are the x-intercepts of this parabola. Visualizing the solution is powerful because it connects the abstract algebra with a concrete image. Graphing the equations provides a clear picture of the intersections and roots, enhancing our understanding of the solutions. This visual confirmation reinforces our algebraic steps, demonstrating how the roots correspond to specific points on the graph, making the solution process more intuitive and memorable.

Why Graphing Matters

So, why bother with graphing in the first place? Well, graphing provides a visual representation of the equation, making it easier to understand the solutions. Instead of just manipulating numbers and symbols, we can see where the graphs intersect or where a graph crosses the x-axis. This visual approach can be especially helpful for those who are visual learners (like me!). Plus, graphing can help us quickly identify the number of real roots an equation has. For example, a graph that doesn't cross the x-axis has no real roots. In more complex equations, graphing can be a lifesaver. Sometimes, algebraic solutions can be very complicated or even impossible to find analytically. In these cases, graphing or using numerical methods to approximate the roots is the way to go. Graphing matters because it bridges the gap between algebra and geometry, offering a comprehensive way to solve equations. It transforms abstract equations into visual stories, making solutions more intuitive and accessible. This method is invaluable for its ability to reveal the number and approximate values of roots, especially in situations where algebraic methods fall short, highlighting the power of visualization in problem-solving.

Choosing the Correct Answer

Alright, we've done the hard work of finding the roots. Now, let's look at the answer choices provided. We found that the roots of the polynomial equation are x=βˆ’3x = -3 and x=4x = 4. Looking at the options, we see:

  • A. -12, 12
  • B. -4, 3

Neither of these options matches our roots. There seems to be a slight mix-up in the signs, but the values are close! Let's double-check our work to make sure we didn't make any mistakes. Double-checking is super important in math! We simplified the original equation to x2βˆ’xβˆ’12=0x^2 - x - 12 = 0, factored it to (xβˆ’4)(x+3)=0(x - 4)(x + 3) = 0, and found the roots x=4x = 4 and x=βˆ’3x = -3. Everything looks good! It seems there might be a typo in the answer choices. The correct roots are -3 and 4, which are not listed in the options provided. Choosing the correct answer emphasizes the importance of matching our solutions with the given options, but also highlights the need for critical thinking when discrepancies arise. Double-checking our work reassures us of our accuracy and prompts us to consider the possibility of errors in the provided choices. This step ensures we confidently identify the true roots, even when faced with imperfect answer options, reinforcing our problem-solving skills and attention to detail.

Conclusion

So, there you have it, guys! We successfully found the roots of the polynomial equation by simplifying, factoring, and understanding the graphical representation. Even though the answer choices weren't a perfect match, we were able to confidently determine the correct roots: -3 and 4. Remember, the key to solving these kinds of problems is to break them down into smaller, manageable steps. Transforming the equation, factoring, finding the roots, and visualizing the solution are all important pieces of the puzzle. And always, always double-check your work! Keep up the awesome work, and happy problem-solving! In conclusion, mastering the process of finding roots through simplification, factoring, and graphical understanding empowers us to tackle polynomial equations with confidence. This comprehensive approach, combined with diligent double-checking, ensures accuracy and reinforces our problem-solving abilities. By embracing these strategies, we unlock the solutions and gain a deeper appreciation for the interconnectedness of algebra and geometry, fostering a resilient and successful approach to mathematical challenges.