Solving X^4 + 6x^2 + 5 = 0 With U-Substitution A Step-by-Step Guide
Hey there, math enthusiasts! Ever stumbled upon an equation that looks a bit intimidating at first glance? Well, today we're going to tackle one of those together. We'll be diving into the equation x⁴ + 6x² + 5 = 0 and cracking it using a clever technique called u-substitution. Trust me, it's not as scary as it sounds! We'll break it down step by step, so by the end of this article, you'll be a pro at solving equations like this.
Understanding the Problem
Before we jump into the solution, let's make sure we understand what we're dealing with. The equation x⁴ + 6x² + 5 = 0 is a quartic equation, which means it's a polynomial equation where the highest power of the variable x is 4. Now, quartic equations can sometimes look daunting, but this one has a special form that makes it easier to handle. Notice that the powers of x are even (4 and 2). This is a clue that we can use a substitution to simplify the equation and transform it into a more familiar form.
Think of it like this: we're trying to find the values of x that, when plugged into the equation, make the whole thing equal to zero. These values are called the roots or solutions of the equation. Our mission is to find all the roots of this quartic equation. So, let's get started and see how u-substitution can help us achieve this goal!
The Magic of u-Substitution
Here comes the exciting part! U-substitution is a technique that allows us to simplify complex equations by replacing a part of the equation with a single variable, usually u. In our case, the key observation is that x⁴ can be written as (x²)². This suggests that we can substitute u for x², which will transform our quartic equation into a quadratic equation—something we're much more comfortable dealing with. Let's do it!
Let u = x². Now, we can rewrite our equation x⁴ + 6x² + 5 = 0 in terms of u. Replacing x² with u, we get:
u² + 6u + 5 = 0
See? It looks much friendlier already! This is a quadratic equation in u, and we know how to solve these. We've effectively transformed a tricky quartic equation into a simpler quadratic equation using u-substitution. Now, let's solve this quadratic equation for u.
Solving the Quadratic Equation
Now that we have our quadratic equation u² + 6u + 5 = 0, we need to find the values of u that satisfy this equation. There are several ways to solve quadratic equations, but one of the most common methods is factoring. We're looking for two numbers that multiply to 5 (the constant term) and add up to 6 (the coefficient of the u term). Can you think of what those numbers might be?
The numbers are 1 and 5! So, we can factor the quadratic equation as follows:
(u + 1)(u + 5) = 0
Now, for this product to be equal to zero, at least one of the factors must be zero. This gives us two possible solutions for u:
u + 1 = 0 or u + 5 = 0
Solving these simple equations, we get:
u = -1 or u = -5
Great! We've found the values of u that satisfy our quadratic equation. But remember, we're not interested in u itself; we want to find the values of x. So, we need to go back to our original substitution and solve for x.
Back to x: Finding the Solutions
We're almost there! Remember that we made the substitution u = x². Now, we need to substitute back the values we found for u and solve for x. We have two cases to consider:
Case 1: u = -1
Substituting u = -1 into our substitution equation, we get:
x² = -1
To solve for x, we take the square root of both sides:
x = ±√(-1)
Now, remember that the square root of -1 is defined as the imaginary unit, denoted by i. So, we have:
x = ±i
These are two of our solutions: x = i and x = -i. These are complex solutions since they involve the imaginary unit i.
Case 2: u = -5
Similarly, substituting u = -5 into u = x², we get:
x² = -5
Taking the square root of both sides:
x = ±√(-5)
We can rewrite √(-5) as √(5 * -1) = √(5) * √(-1) = √5 * i. So, we have:
x = ±i√5
These are our other two solutions: x = i√5 and x = -i√5. These are also complex solutions.
The Grand Finale: Our Solutions
We've done it! We've successfully solved the equation x⁴ + 6x² + 5 = 0 using u-substitution. Let's recap our solutions. We found four values of x that satisfy the equation:
- x = i
- x = -i
- x = i√5
- x = -i√5
So, the solutions to the equation are x = ±i and x = ±i√5. This corresponds to option B in the original problem. You nailed it!
Why This Works: The Big Picture
Let's take a step back and appreciate what we've accomplished. We started with a quartic equation that looked a bit scary. But by using the clever technique of u-substitution, we transformed it into a quadratic equation, which we know how to solve. This highlights a powerful problem-solving strategy in mathematics: transforming a complex problem into a simpler one. By recognizing the special form of the quartic equation (even powers of x), we were able to make a substitution that simplified the equation.
U-substitution is a versatile tool that can be used in many different contexts in mathematics, especially in calculus when dealing with integrals. The key idea is to identify a part of the expression that, when replaced with a new variable, simplifies the overall expression. It's like finding a hidden lever that, when pulled, makes the whole machine work more smoothly. Remember guys, it is all about simplifying and making it manageable.
Alternative Approaches (Just for Fun!)
While u-substitution was the star of the show today, it's worth noting that there might be other ways to solve this equation. For instance, one could try factoring the quartic expression directly. However, u-substitution often provides a more straightforward and systematic approach, especially when the equation has this specific form with even powers of x.
Also, for those familiar with complex numbers, another perspective is to recognize that quartic equations can have up to four complex roots. Our solutions, x = ±i and x = ±i√5, confirm this fact. Each solution represents a point in the complex plane, and understanding complex numbers provides a deeper insight into the nature of these solutions.
Practice Makes Perfect
Now that you've seen how to solve this type of equation using u-substitution, the best way to master the technique is to practice! Try solving similar equations. For example, you could try:
- x⁴ - 5x² + 4 = 0
- x⁴ + 8x² + 15 = 0
- 2x⁴ + 7x² + 3 = 0
Remember, the key is to identify the appropriate substitution (u = x² in these cases) and then solve the resulting quadratic equation. Don't be afraid to make mistakes—that's how we learn! The more you practice, the more comfortable you'll become with u-substitution and the better you'll be at solving these types of equations.
Conclusion: You've Got This!
Congratulations! You've successfully navigated the world of quartic equations and mastered the art of u-substitution. We took a seemingly complex problem and broke it down into manageable steps. We learned how to recognize the special form of the equation, make the appropriate substitution, solve the resulting quadratic equation, and then substitute back to find the solutions for the original variable. This is a powerful skill that will serve you well in your mathematical journey.
So, the next time you encounter a tricky equation, remember the magic of u-substitution. Think of it as a tool in your mathematical toolbox that you can pull out whenever you need to simplify a problem. Keep practicing, keep exploring, and most importantly, keep enjoying the beauty and power of mathematics!
