Solving X^4 + 5x^2 = 50 A Step-by-Step Guide

Hey guys! Let's dive into solving a fascinating biquadratic equation. We're tackling the equation x^4 + 5x^2 = 50. This might look intimidating at first, but don't worry, we'll break it down step by step and make it super easy to understand. Solving biquadratic equations like this one involves a clever trick: recognizing the quadratic form. A biquadratic equation is essentially a quadratic equation in disguise. When you see terms with powers of x that are multiples of 2 (like x⁴ and x²), that’s your cue that a substitution might be helpful. The key to conquering these equations lies in recognizing their hidden quadratic nature. By employing a simple substitution, we can transform the biquadratic into a familiar quadratic equation. Once we've solved the quadratic, we backtrack to find the solutions for the original variable, x. This process not only simplifies the equation but also allows us to apply well-known methods like factoring or the quadratic formula. So, buckle up, and let's unlock the secrets of biquadratic equations together! We’ll cover everything from the initial setup to the final solutions, ensuring you’ve got a solid grasp of the process.

1. Transforming the Equation into Quadratic Form

So, the first thing we need to do is get this equation into a form we can actually work with. Right now, it looks a bit scary with that x to the fourth power. But here’s the magic trick: we can make a simple substitution to turn it into something much friendlier – a good old quadratic equation! Let’s think about what we have: x^4 + 5x^2 = 50. Notice that x⁴ is just (x²)². This is our key insight. We're going to introduce a new variable, let's call it y, and we'll say that y = x². This is where the magic happens. By making this substitution, we’re essentially reframing the equation in terms of y. It allows us to see the underlying quadratic structure that was hidden beneath the fourth power. So, instead of dealing with x⁴, we can now work with y², which is much more manageable. When we replace x² with y, then x⁴ becomes y². Now, our equation transforms beautifully! We started with x^4 + 5x^2 = 50, and after the substitution, we get y^2 + 5y = 50. See? Much better! But we're not quite ready to solve just yet. Remember, quadratic equations are happiest when they're set equal to zero. So, our next step is to move that 50 over to the left side. Subtracting 50 from both sides gives us y^2 + 5y - 50 = 0. Ta-da! We've done it. We've successfully transformed our biquadratic equation into a standard quadratic equation. This is a crucial step because we now have a form that we know how to solve. We've turned a complicated problem into a familiar one, and that's always a good feeling. Now we have a quadratic equation in the form of ay² + by + c = 0, where a = 1, b = 5, and c = -50. This form is perfect for applying methods like factoring, completing the square, or the quadratic formula. In the next section, we’ll explore how to solve this quadratic equation and find the values of y. Once we have the values of y, we’ll then backtrack to find the values of x that satisfy our original biquadratic equation. So, hang tight, we're making great progress! This transformation is the heart of solving biquadratic equations, and you've nailed it. Keep up the awesome work!

2. Solving the Quadratic Equation

Alright, guys, we've successfully transformed our biquadratic equation into a quadratic equation: y^2 + 5y - 50 = 0. Now comes the fun part – actually solving it! There are a couple of ways we can tackle this, but let's start with factoring. Factoring is often the quickest route if we can find two numbers that play nicely together. Remember, when we factor a quadratic, we're looking for two binomials that multiply to give us our equation. In this case, we need two numbers that multiply to -50 and add up to 5. Think about the factors of 50: 1 and 50, 2 and 25, 5 and 10. Bingo! 10 and -5 look promising. 10 multiplied by -5 is -50, and 10 plus -5 is 5. Perfect! So, we can factor our quadratic equation as (y + 10)(y - 5) = 0. See how those numbers fit right in? The +10 and -5 are the key. Now, here’s the cool part about factoring: if the product of two things is zero, then at least one of them has to be zero. It’s a fundamental principle that helps us solve for y. So, either (y + 10) = 0 or (y - 5) = 0. Let's solve each of these little equations separately. If y + 10 = 0, then subtracting 10 from both sides gives us y = -10. And if y - 5 = 0, then adding 5 to both sides gives us y = 5. Awesome! We've found our two solutions for y: y = -10 and y = 5. We're halfway there! But remember, we're not actually trying to solve for y. We want to find the values of x that satisfy our original equation. So, we need to take these values of y and plug them back into our substitution, which was y = x². This is where we bring it all full circle. We’ll take each value of y and use it to create a new equation in terms of x. Then, we'll solve those equations to find our final answers. It's like we're unwrapping a present, layer by layer. We started with a tricky biquadratic, transformed it into a quadratic, solved for y, and now we're ready to find x. Each step has brought us closer to the solution, and you’re doing fantastic! In the next section, we’ll tackle the back-substitution and find those elusive values of x. Get ready to see the final pieces of the puzzle fall into place!

3. Back-Substitution to Find the Values of x

Okay, we've done the heavy lifting and found the values of y: y = -10 and y = 5. Now it's time to bring it all home and find the values of x. Remember our original substitution: y = x². This is our key to unlocking the solutions for x. We're going to take each value of y and plug it back into this equation. Let's start with y = -10. If y = x², then we have x^2 = -10. Now we need to solve for x. To do that, we take the square root of both sides. But here's a little twist: the square root of a negative number involves imaginary numbers! This is where the complex solutions come into play. When we take the square root of both sides, we get x = ±√(-10). Remember, we always include both the positive and negative square roots. Now, we can simplify this a bit using the imaginary unit, i, where i = √(-1). So, √(-10) can be written as √(10) * √(-1), which is √(10) * i. Therefore, our solutions for this case are x = ±√(10)i. These are two of our four solutions, and they're complex numbers. Don't let that scare you; they're just as valid as real number solutions! Now, let's tackle the other value of y, which is y = 5. If y = x², then we have x^2 = 5. Again, we take the square root of both sides to solve for x. This gives us x = ±√5. These are our other two solutions, and they're real numbers. So, we have x = √5 and x = -√5. Fantastic! We've found all four solutions for x. We have two real solutions and two complex solutions. This is exactly what we expect from a biquadratic equation, which is essentially a fourth-degree polynomial equation. Fourth-degree polynomials have four roots, counting multiplicity, and we've found them all. To recap, our solutions are: x = √(10)i, x = -√(10)i, x = √5, and x = -√5. We’ve successfully navigated the world of biquadratic equations! We transformed it, factored it, back-substituted, and found all the solutions. You’ve shown some serious problem-solving skills, guys! Remember, the key to these problems is recognizing the hidden quadratic form and using substitution to simplify the equation. In the final section, we’ll bring it all together and present our final answer in a clear and concise way.

4. Presenting the Final Solutions

Alright, let's bring it all together and present our final solutions in a clear and organized way. We've been on quite a journey, guys, transforming, solving, and back-substituting, and now we're at the finish line! We started with the equation x^4 + 5x^2 = 50, and after all our hard work, we've discovered the four values of x that make this equation true. Remember, we found two real solutions and two complex solutions. Our real solutions were x = √5 and x = -√5. These are the values of x that, when squared, give us 5. And our complex solutions were x = √(10)i and x = -√(10)i. These solutions involve the imaginary unit i, which is the square root of -1. When we square these values, we get -10, which fits perfectly into our transformed equation. So, to present our final answer in a neat format, we can write: The solutions are x = √5, -√5, √(10)i, -√(10)i. There you have it! We've successfully solved the biquadratic equation. This is a fantastic achievement, and you should be proud of your ability to tackle such a problem. We've demonstrated a powerful problem-solving technique: breaking down a complex problem into smaller, more manageable steps. We used substitution to transform the equation, factoring to solve the quadratic form, and back-substitution to find the original solutions. This approach can be applied to many other types of mathematical problems, so keep it in your toolkit! Solving biquadratic equations might seem daunting at first, but with a systematic approach and a little bit of algebraic manipulation, they become much more approachable. Remember the key steps: identify the quadratic form, make a substitution, solve the resulting quadratic equation, and then back-substitute to find the solutions for the original variable. And most importantly, don't be afraid of complex numbers! They're just another type of number, and they play an important role in many areas of mathematics and science. So, congratulations on mastering this biquadratic equation! You've added a valuable skill to your mathematical arsenal. Keep practicing and exploring, and you'll continue to grow your problem-solving abilities. Great job, everyone!

The solution(s) is/are x = √5, -√5, √(10)i, -√(10)i.