Solving The Quartic Equation: X⁴ - 13x² + 36 = 0

Hey guys! Today, we're diving into the fascinating world of equations, specifically a quartic equation. Don't let the fancy name scare you; it's just a polynomial equation where the highest power of the variable (in our case, 'x') is 4. We're going to break down how to solve the equation x⁴ - 13x² + 36 = 0. This type of equation might seem intimidating at first glance, but with a clever substitution, we can transform it into a quadratic equation, which we know how to handle. So, buckle up, grab your pencils, and let's get started!

Understanding Quartic Equations

Before we jump into the solution, let's understand what makes this equation a quartic equation. A quartic equation is a polynomial equation of the fourth degree. The general form of a quartic equation is ax⁴ + bx³ + cx² + dx + e = 0, where 'a', 'b', 'c', 'd', and 'e' are constants, and 'a' is not equal to zero. Our equation, x⁴ - 13x² + 36 = 0, fits this form, where a = 1, b = 0, c = -13, d = 0, and e = 36. Notice that the terms with x³ and x are missing, which simplifies our equation a bit. Quartic equations can have up to four solutions (also called roots), which can be real or complex numbers. Finding these solutions is the goal of solving the equation. There are several methods to solve quartic equations, including factoring, using the quartic formula (which is quite complex), and employing substitution techniques, which is the method we'll use here. Understanding the nature of quartic equations and their possible solutions helps us approach the problem with the right mindset. We're essentially looking for the values of 'x' that make the equation true, and knowing that there could be up to four such values gives us a target to aim for. Remember, math is like a puzzle, and understanding the pieces is the first step to solving it!

The Substitution Trick

The key to solving this particular quartic equation lies in a clever substitution. We can notice that the equation x⁴ - 13x² + 36 = 0 has a special structure: it only contains even powers of 'x'. This means we can rewrite x⁴ as (x²)² and make a substitution to simplify the equation. Let's introduce a new variable, say 'y', such that y = x². By substituting 'y' for x², we transform our quartic equation into a quadratic equation in terms of 'y'. This is a crucial step because we have well-established methods for solving quadratic equations. When we substitute y = x² into x⁴ - 13x² + 36 = 0, we get (x²)² - 13(x²) + 36 = 0, which then becomes y² - 13y + 36 = 0. Now we have a standard quadratic equation that we can solve using factoring, the quadratic formula, or completing the square. This substitution technique is a powerful tool in mathematics, allowing us to simplify complex problems by changing the variables. It's like using a different lens to view the problem, revealing a simpler structure underneath. The beauty of this method is that it transforms a seemingly difficult quartic equation into a manageable quadratic equation. This highlights the importance of recognizing patterns and structures in mathematical problems, as they often lead to elegant solutions. So, remember this trick: when you see an equation with even powers, consider making a substitution to simplify it!

Solving the Quadratic Equation

Now that we've transformed our quartic equation into the quadratic equation y² - 13y + 36 = 0, let's solve for 'y'. There are a couple of ways we can tackle this: factoring or using the quadratic formula. Factoring is often the quickest method if we can easily find two numbers that multiply to 36 and add up to -13. In this case, those numbers are -4 and -9, since (-4) * (-9) = 36 and (-4) + (-9) = -13. Therefore, we can factor the quadratic equation as (y - 4)(y - 9) = 0. This factored form tells us that the equation is satisfied if either (y - 4) = 0 or (y - 9) = 0. Solving these two simple equations gives us the solutions for 'y': y = 4 and y = 9. Alternatively, if factoring doesn't come easily, we can always use the quadratic formula. The quadratic formula states that for an equation of the form ay² + by + c = 0, the solutions for 'y' are given by y = (-b ± √(b² - 4ac)) / (2a). In our case, a = 1, b = -13, and c = 36. Plugging these values into the quadratic formula, we get y = (13 ± √((-13)² - 4 * 1 * 36)) / (2 * 1) = (13 ± √(169 - 144)) / 2 = (13 ± √25) / 2 = (13 ± 5) / 2. This gives us two solutions: y = (13 + 5) / 2 = 9 and y = (13 - 5) / 2 = 4, which are the same solutions we found by factoring. So, we've successfully solved for 'y', but remember, we're ultimately trying to find the values of 'x'. We're not done yet!

Back to 'x': Finding the Solutions

We've found the values of 'y', but we need to remember that we made a substitution: y = x². To find the values of 'x', we need to reverse this substitution. We have two solutions for 'y': y = 4 and y = 9. Let's take each one and substitute it back into the equation y = x². First, let's consider y = 4. Substituting this in, we get x² = 4. To solve for 'x', we take the square root of both sides, remembering that we need to consider both the positive and negative square roots. So, x = ±√4, which gives us two solutions: x = 2 and x = -2. Next, let's consider y = 9. Substituting this into y = x², we get x² = 9. Again, we take the square root of both sides, considering both positive and negative roots. This gives us x = ±√9, which leads to two more solutions: x = 3 and x = -3. Therefore, the equation x⁴ - 13x² + 36 = 0 has four solutions: x = 2, x = -2, x = 3, and x = -3. These are the values of 'x' that make the original equation true. It's important to remember to go back to the original variable after solving for the substituted variable. This step ensures that we answer the question that was initially asked. And there you have it! We've successfully found all the solutions to the quartic equation.

Verification and Conclusion

To ensure we haven't made any mistakes, it's always a good idea to verify our solutions. We can do this by plugging each of our solutions (x = 2, x = -2, x = 3, and x = -3) back into the original equation x⁴ - 13x² + 36 = 0 and checking if the equation holds true. Let's start with x = 2: (2)⁴ - 13(2)² + 36 = 16 - 13(4) + 36 = 16 - 52 + 36 = 0. The equation holds true for x = 2. Now let's check x = -2: (-2)⁴ - 13(-2)² + 36 = 16 - 13(4) + 36 = 16 - 52 + 36 = 0. It also holds true for x = -2. Next, let's try x = 3: (3)⁴ - 13(3)² + 36 = 81 - 13(9) + 36 = 81 - 117 + 36 = 0. The equation is satisfied for x = 3. Finally, let's check x = -3: (-3)⁴ - 13(-3)² + 36 = 81 - 13(9) + 36 = 81 - 117 + 36 = 0. It also holds true for x = -3. Since all four solutions satisfy the original equation, we can confidently say that we've found all the correct solutions. In conclusion, we've successfully solved the quartic equation x⁴ - 13x² + 36 = 0 by using a clever substitution to transform it into a quadratic equation. We solved the quadratic equation and then reversed the substitution to find the solutions for 'x'. The solutions are x = 2, x = -2, x = 3, and x = -3. Remember, guys, the key to solving complex problems often lies in finding the right strategy to simplify them. Keep practicing, and you'll become a math whiz in no time!