Solving Quadratic Equations: Find X In 6x² - X - 15 = 0

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Hey guys! Today, we're diving into the exciting world of quadratic equations. We're going to tackle the equation 6x² - x - 15 = 0 and find the values of x that make this equation true. Don't worry; it's not as scary as it looks! We'll break it down step by step, so you'll be solving quadratic equations like a pro in no time. So, let’s get started and see how we can crack this mathematical puzzle together!

Understanding Quadratic Equations

Before we jump into solving our specific equation, let's make sure we're all on the same page about what a quadratic equation actually is. In simple terms, a quadratic equation is a polynomial equation of the second degree. That might sound like a mouthful, but all it means is that the highest power of the variable (in our case, 'x') is 2. The general form of a quadratic equation is ax² + bx + c = 0, where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero.

Why is 'a' not equal to zero important? Well, if 'a' were zero, the x² term would disappear, and we'd be left with a linear equation instead of a quadratic one. Linear equations are a whole different ball game! The coefficients a, b, and c play crucial roles in determining the shape and position of the parabola when the quadratic equation is graphed, and they also influence the solutions, or roots, of the equation.

Now, back to our equation: 6x² - x - 15 = 0. Can you see how it fits the general form? Here, 'a' is 6, 'b' is -1 (remember, the minus sign is important!), and 'c' is -15. Identifying these coefficients is the first step in choosing the right method to solve the equation. Understanding the structure of a quadratic equation is fundamental because it dictates the methods we can use to find the solutions. Quadratic equations pop up everywhere in math and science, from calculating the trajectory of a ball to designing bridges, so mastering them is super useful.

Methods to Solve Quadratic Equations

Okay, now that we know what a quadratic equation is, let's explore the different ways we can solve them. There are three main methods in our toolkit: factoring, completing the square, and using the quadratic formula. Each method has its strengths and weaknesses, and the best one to use often depends on the specific equation you're dealing with.

  • Factoring: This method involves breaking down the quadratic expression into two binomials. It's like reverse-distributing! Factoring is often the quickest method when it works, but it's not always easy to spot the factors, especially when the coefficients are large or the equation doesn't have simple integer solutions.
  • Completing the Square: This method involves manipulating the equation to create a perfect square trinomial on one side. It's a bit more involved than factoring, but it's a powerful technique that always works. Plus, understanding completing the square helps you understand the derivation of the quadratic formula.
  • Quadratic Formula: This is the ultimate Swiss Army knife of quadratic equation solving. It works for any quadratic equation, no matter how messy the coefficients are. The quadratic formula is: x = (-b ± √(b² - 4ac)) / 2a. It might look intimidating, but once you get the hang of plugging in the values, it's pretty straightforward. Understanding the discriminant (b² - 4ac) within the formula helps predict the nature of the solutions, whether they are real, distinct, or complex.

For our equation, 6x² - x - 15 = 0, we could use any of these methods, but let's start by trying factoring since it's often the fastest if it works. Factoring is like finding the puzzle pieces that fit together to form the quadratic equation, making it a valuable skill in algebra. Each method provides a unique approach to solving quadratics, ensuring that there's a strategy available no matter the equation's complexity.

Solving 6x² - x - 15 = 0 by Factoring

Alright, let’s roll up our sleeves and tackle 6x² - x - 15 = 0 using the factoring method. Factoring might sound intimidating, but it's like solving a puzzle, and once you get the hang of it, it can be super satisfying. The main idea behind factoring is to rewrite the quadratic expression as a product of two binomials. In other words, we want to find two expressions that, when multiplied together, give us our original equation.

So, we're looking for something that looks like this: (Ax + B)(Cx + D) = 6x² - x - 15. Where A, B, C, and D are constants that we need to figure out. The trick is to think about how the coefficients in the binomials will multiply to give us the coefficients in our quadratic equation.

First, let's focus on the 6x² term. We need two terms that multiply to give us 6x². Possible pairs are (6x and x) or (3x and 2x). Let’s try (3x and 2x) for now. Next, let's look at the constant term, -15. We need two numbers that multiply to give us -15. Possible pairs are (-5 and 3), (5 and -3), (-1 and 15), or (1 and -15). Now, we need to play around with these pairs to see which combination will give us the correct middle term, which is -x. This is where the trial and error comes in, but don't worry, it's all part of the fun! After some experimenting, we find that (3x - 5)(2x + 3) works perfectly. Let's check it: (3x - 5)(2x + 3) = 6x² + 9x - 10x - 15 = 6x² - x - 15. Bingo!

Now that we've factored the equation, we have (3x - 5)(2x + 3) = 0. For this product to be zero, at least one of the factors must be zero. So, we set each factor equal to zero and solve for x:

  • 3x - 5 = 0
  • 2x + 3 = 0

These linear equations are much easier to solve. Solving these mini-equations gives us the solutions to the original quadratic equation. Factoring simplifies the problem by breaking it into manageable parts, showcasing the power of algebraic manipulation.

Finding the Solutions for x

Great job, guys! We've successfully factored the quadratic equation 6x² - x - 15 = 0 into (3x - 5)(2x + 3) = 0. Now comes the final step: finding the values of 'x' that make this equation true. Remember, the whole idea behind factoring is that if the product of two factors is zero, then at least one of those factors must be zero. This is a fundamental principle in algebra, and it's what allows us to solve for x once we have the factored form.

So, we have two possibilities to consider:

  1. The first factor, 3x - 5, is equal to zero.
  2. The second factor, 2x + 3, is equal to zero.

Let's take each case one at a time and solve for 'x'.

Case 1: 3x - 5 = 0

To solve for 'x', we need to isolate it on one side of the equation. We can do this by adding 5 to both sides: 3x - 5 + 5 = 0 + 5, which simplifies to 3x = 5. Now, to get 'x' by itself, we divide both sides by 3: 3x / 3 = 5 / 3, which gives us x = 5/3. So, one of our solutions is x = 5/3. This step demonstrates the basic algebraic principle of maintaining equality by performing the same operation on both sides.

Case 2: 2x + 3 = 0

We follow the same process here. First, subtract 3 from both sides: 2x + 3 - 3 = 0 - 3, which simplifies to 2x = -3. Then, divide both sides by 2: 2x / 2 = -3 / 2, which gives us x = -3/2. So, our second solution is x = -3/2. This mirrors the process in Case 1, reinforcing the methodical approach to solving linear equations.

Therefore, the solutions to the quadratic equation 6x² - x - 15 = 0 are x = 5/3 and x = -3/2. We've done it! We've successfully navigated through the factoring process and found the two values of 'x' that satisfy the equation. Each solution represents a point where the parabola intersects the x-axis, linking algebra to graphical representation.

Expressing the Solutions

Now that we've found our two solutions, x = 5/3 and x = -3/2, let's make sure we express them in the way the problem asked us to. The instructions said to write both solutions separated by a comma. So, that's exactly what we'll do! It's super important to pay attention to the instructions in math problems. Sometimes, you can do all the hard work of solving the equation but lose points if you don't present the answer in the correct format.

So, our final answer, written as requested, is: 5/3, -3/2. There you have it! We've not only solved the quadratic equation but also made sure we communicated our answer clearly and correctly. Presenting solutions in the specified format ensures clear communication and adherence to mathematical conventions.

Alternative Methods: Quadratic Formula

Just to show you how versatile quadratic equations are, let's quickly take a peek at how we could have solved 6x² - x - 15 = 0 using the quadratic formula. Remember, the quadratic formula is our trusty backup that works for any quadratic equation, even when factoring is tricky. The quadratic formula is:

x = (-b ± √(b² - 4ac)) / 2a

Where 'a', 'b', and 'c' are the coefficients from our equation ax² + bx + c = 0. In our case, a = 6, b = -1, and c = -15. Now, we just plug these values into the formula:

x = (-(-1) ± √((-1)² - 4 * 6 * -15)) / (2 * 6)

Let's simplify this step by step:

  • First, simplify the negative signs: x = (1 ± √(1 + 360)) / 12
  • Then, add the numbers under the square root: x = (1 ± √361) / 12
  • The square root of 361 is 19: x = (1 ± 19) / 12

Now we have two possibilities, one with the plus sign and one with the minus sign:

  • x = (1 + 19) / 12 = 20 / 12 = 5/3
  • x = (1 - 19) / 12 = -18 / 12 = -3/2

See? We got the same solutions as when we factored! This illustrates the power and reliability of the quadratic formula, offering a guaranteed solution path regardless of the equation's factorability.

Conclusion

Awesome job, guys! We successfully solved the quadratic equation 6x² - x - 15 = 0 using factoring and even double-checked our answers with the quadratic formula. We found that the solutions are x = 5/3 and x = -3/2. Remember, practice makes perfect, so keep solving those quadratic equations, and you'll become a master in no time! Whether factoring, completing the square, or applying the quadratic formula, each method provides a pathway to solving these fundamental equations. Keep up the great work, and you'll conquer any math challenge that comes your way! Understanding and solving quadratic equations unlocks a deeper understanding of algebra and its applications in various fields.