Solve Quadratic Equations: A Step-by-Step Guide

Hey math enthusiasts! Let's dive into the world of solving quadratic equations. Specifically, we're going to tackle the equation: $\frac{x^2-19}{x^2-3 x-40}=1+\frac{9}{x-8}$. This might look a little intimidating at first, but trust me, we'll break it down step-by-step to make it super easy to understand. We'll be using a combination of algebraic manipulation and critical thinking to find the solution set. So, buckle up, grab your pencils, and let's get started!

Decoding the Equation: Initial Steps

Alright, guys, our first step is to get rid of those pesky fractions. To do this, we need to find a common denominator. Notice that the denominator on the left side is $x^2 - 3x - 40$. We can factor this to $(x - 8)(x + 5)$. Also, we have $(x - 8)$ in the denominator on the right side. So, the least common denominator (LCD) for this equation is $(x - 8)(x + 5)$. We need to be super careful here, though. Before we start multiplying, let's make sure we understand the domain of our equation. The original equation has denominators of $x^2 - 3x - 40$ and $x - 8$. Therefore, $x$ cannot be equal to $8$ or $-5$, because division by zero is undefined! Keep these restrictions in mind, because if we get one of these as an answer, we'll have to throw it out. Any solution needs to satisfy these conditions. Now, let's multiply both sides of our equation by the LCD, $(x - 8)(x + 5)$:

$(x - 8)(x + 5) * \frac{x^2-19}{x^2-3 x-40} = (x - 8)(x + 5) * (1 + \frac{9}{x-8})$

This will help us eliminate the fractions and simplify the equation. This is the crucial first step in tackling these types of problems. Remember, the goal is always to make the equation easier to work with, and getting rid of fractions is a big win. Keep the domain restrictions in mind throughout the simplification process.

Simplifying the Equation and Avoiding Pitfalls

Now, let's simplify! On the left side of the equation, the $(x^2 - 3x - 40)$ cancels with $(x - 8)(x + 5)$, leaving us with $(x^2 - 19)$. On the right side, we distribute $(x - 8)(x + 5)$ across the terms in the parentheses:

$(x - 8)(x + 5) * \frac{x^2-19}{(x - 8)(x + 5)} = (x - 8)(x + 5) * 1 + (x - 8)(x + 5) * \frac{9}{x-8}$

This simplifies to:

$x^2 - 19 = (x - 8)(x + 5) + 9(x + 5)$

Next, expand the right side of the equation:

$x^2 - 19 = x^2 - 3x - 40 + 9x + 45$

Notice that the $x^2$ terms on both sides seem to cancel out, which simplifies things significantly. Combine the $x$ terms and constant terms: $x^2 - 19 = x^2 + 6x + 5$

Further simplifying the equation becomes:

$0 = 6x + 24$

Remember, we are aiming to get a simplified form that allows us to find the value(s) of $x$ that satisfy the original equation. Make sure you don't lose track of your goal. The more you practice, the faster and more comfortable you'll become with this. Always double-check your work to avoid silly mistakes. Be extremely careful when doing the distribution and simplification steps as this is where errors can easily be made!

Isolating 'x' and Finding Potential Solutions

Now that we've simplified the equation, let's isolate $x$. We have $0 = 6x + 24$. Subtract $24$ from both sides: $-24 = 6x$. Then, divide both sides by $6$: $x = -4$. So, we have found a potential solution, $x = -4$. But, remember those domain restrictions we talked about at the beginning? We have to make sure that $x = -4$ doesn't violate any of them. Since neither $8$ nor $-5$ are equal to $-4$, this is a valid solution! Keep going, we are almost done!

The Final Verification and Solution Set

To be absolutely sure, it's always a good idea to plug our potential solution back into the original equation to verify it. Substitute $x = -4$ into the original equation:

$\frac{(-4)^2-19}{(-4)^2-3 (-4)-40}=1+\frac{9}{-4-8}$

This simplifies to:

$\frac{16-19}{16+12-40}=1+\frac{9}{-12}$

$\frac{-3}{-12} = 1 - \frac{3}{4}$

$\frac{1}{4} = \frac{1}{4}$

It checks out! Therefore, our solution is correct. Thus, the solution set is $\{-4\}$. It's really that straightforward once you get the hang of it! Remember to always check your answers to make sure they are valid, especially in equations involving fractions. You got this, guys!

Summary of Steps and Key Takeaways

Let's recap the steps we took to solve this equation:

1. Identify Domain Restrictions: Determine any values of $x$ that would make the denominators zero. These are the values $x$ cannot be. This crucial step prevents us from making mistakes. These restrictions stay in place throughout the whole problem.
2. Find the LCD: Determine the least common denominator to eliminate fractions.
3. Multiply by the LCD: Multiply both sides of the equation by the LCD. This clears out those pesky fractions, simplifying the equation considerably.
4. Simplify and Expand: Simplify the resulting equation, combining like terms and expanding any expressions.
5. Isolate x: Solve for $x$ by isolating it on one side of the equation.
6. Check for Validity: Make sure that your solutions do not violate any of the domain restrictions you identified in step 1. If a solution is restricted, it must be thrown out.
7. Verify the Solution: Plug the potential solution back into the original equation to check your work. This helps to catch any errors and ensures your answer is correct. This is important!

This methodical approach will help you conquer any quadratic equation you encounter! Always take your time, double-check your work, and don't be afraid to ask for help if you need it. Now go out there and solve some equations!

A. The solution set is $\{-4\}$.

B. The solution set is $\varnothing$. (This is incorrect as we did find a solution, but this is a common answer option to include in the question).