Sum Of Roots: Solving A Complex Equation | Algebra

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Hey guys! Let's dive into a fascinating algebraic problem today. We're going to tackle a complex equation and find the sum of its roots. This type of problem often appears in algebra discussions, and understanding how to solve it can really boost your math skills. So, grab your pencils, and let’s get started!

The Challenge: Deciphering the Equation

Our main goal here is to figure out the sum of the roots for the given equation:

(2x^2 + 10x + 8) / (4x^2 + 22x + 24) = ((4x - 6)^2) / (16x^2 - 36)

At first glance, it looks a bit intimidating, right? But don't worry, we'll break it down step by step. The key to solving complex equations like this is to simplify them first. This involves factoring, canceling out common terms, and rearranging the equation into a more manageable form. We need to simplify each side, identify potential exclusions, and then consolidate it into a solvable quadratic form. Factoring is your friend here; it helps in simplifying and spotting cancellations. So, let’s roll up our sleeves and get into the nitty-gritty details of how to solve this equation.

Step-by-Step Solution

1. Factor Everything

Factoring is the magic word in algebra! It helps us simplify complex expressions. Let's start by factoring both the numerator and the denominator on each side of the equation.

  • Numerator of the left side: 2x^2 + 10x + 8. We can factor out a 2, giving us 2(x^2 + 5x + 4). Then, we factor the quadratic expression: 2(x + 1)(x + 4).
  • Denominator of the left side: 4x^2 + 22x + 24. We can factor out a 2, giving us 2(2x^2 + 11x + 12). Factoring the quadratic expression, we get 2(x + 4)(2x + 3).
  • Numerator of the right side: (4x - 6)^2. This is a square, so we can write it as (4x - 6)(4x - 6). We can factor out a 2 from each term, giving us [2(2x - 3)]^2 which simplifies to 4(2x - 3)^2.
  • Denominator of the right side: 16x^2 - 36. This is a difference of squares, so we can factor it as (4x - 6)(4x + 6). Again, we can factor out constants, giving us 4(2x - 3)(2x + 3).

2. Rewrite the Equation with Factored Forms

Now that we've factored everything, let’s rewrite the original equation using these factored forms:

[2(x + 1)(x + 4)] / [2(x + 4)(2x + 3)] = [4(2x - 3)^2] / [4(2x - 3)(2x + 3)]

This looks a bit cleaner already, doesn't it? Factoring helps us see the common terms that we can cancel out.

3. Cancel Out Common Terms

Next, we’ll cancel out the common factors in the numerators and denominators. This will simplify the equation even further.

  • On the left side, we can cancel out the 2 and the (x + 4).
  • On the right side, we can cancel out the 4 and one of the (2x - 3) terms.

After canceling, our equation becomes:

(x + 1) / (2x + 3) = (2x - 3) / (2x + 3)

See how much simpler it looks now? This is the power of factoring and canceling!

4. Identify Excluded Values

Before we go any further, it’s crucial to identify any excluded values. These are the values of x that would make the denominator zero, which is a big no-no in math because division by zero is undefined.

From the original equation, we have denominators of 4x^2 + 22x + 24 and 16x^2 - 36. Setting each factored denominator to zero helps us find the excluded values.

  • 2(x + 4)(2x + 3) = 0 gives us x = -4 and x = -3/2.
  • 4(2x - 3)(2x + 3) = 0 gives us x = 3/2 and x = -3/2.

So, our excluded values are x = -4, x = -3/2, and x = 3/2. We need to remember these and make sure our solutions aren’t any of these values.

5. Solve the Simplified Equation

Now that we’ve simplified the equation and identified the excluded values, we can solve for x. We start by cross-multiplying:

(x + 1)(2x + 3) = (2x - 3)(2x + 3)

Expand both sides:

2x^2 + 3x + 2x + 3 = 4x^2 - 9

Combine like terms:

2x^2 + 5x + 3 = 4x^2 - 9

Move everything to one side to set the equation to zero:

0 = 2x^2 - 5x - 12

Now we have a quadratic equation to solve. We can solve this by factoring, using the quadratic formula, or completing the square. Let’s try factoring:

0 = (2x + 3)(x - 4)

This gives us two possible solutions:

  • 2x + 3 = 0 => x = -3/2
  • x - 4 = 0 => x = 4

6. Check for Extraneous Solutions

Remember those excluded values we found earlier? We need to check if any of our solutions are on that list. We found that x = -3/2 is an excluded value, so it’s not a valid solution. This is called an extraneous solution.

That leaves us with x = 4 as our only valid solution.

7. Find the Sum of the Roots

Since we only have one valid root, x = 4, the sum of the roots is simply 4. So, the answer is:

Sum of roots = 4

Key Concepts and Techniques

To successfully solve this problem, we used several key concepts and techniques from algebra. Let's recap them to reinforce our understanding:

  • Factoring: This is the backbone of simplifying algebraic expressions. We factored quadratic expressions, differences of squares, and common factors to make the equation more manageable.
  • Canceling Common Terms: After factoring, we canceled out common factors in the numerators and denominators. This is a crucial step in simplifying rational expressions.
  • Identifying Excluded Values: We found the values of x that would make the denominator zero and excluded them from our possible solutions. This ensures we don't divide by zero.
  • Solving Quadratic Equations: We ended up with a quadratic equation, which we solved by factoring. Other methods include using the quadratic formula or completing the square.
  • Checking for Extraneous Solutions: We verified that our solutions were not excluded values, ensuring we only kept the valid ones.

Why is This Important?

Understanding how to solve equations like this is super important for a few reasons:

  • Problem-Solving Skills: It helps you develop problem-solving skills that are useful not just in math, but in many areas of life.
  • Foundation for Advanced Math: It lays the groundwork for more advanced topics in algebra and calculus.
  • Standardized Tests: These types of questions often appear on standardized tests like the SAT and ACT.

Practice Makes Perfect

Solving complex equations can be tricky at first, but the more you practice, the better you’ll get. Try tackling similar problems to build your skills and confidence. Remember, the key is to break the problem down into smaller, manageable steps.

Conclusion

So, guys, we've successfully navigated through a complex algebraic equation and found the sum of its roots. We factored, canceled, solved, and checked our answers. Remember, math isn't about magic; it's about logic and practice. Keep at it, and you’ll become a pro at solving these types of problems!

If you have any questions or want to try another problem, let me know in the comments. Happy solving!