Solving Quadratic Equations: A Step-by-Step Guide
Hey math enthusiasts! Ever stumbled upon an equation that looks a bit intimidating? Well, don't worry, because today, we're diving headfirst into the world of solving quadratic equations with a specific example: $\frac{x2}{\sqrt{x2-c2}}=\frac{c2}{\sqrt{x2-c2}}+39$, where c is a positive constant. Our mission? To uncover one of the solutions to this equation. Sounds like fun, right? Let's break it down step-by-step and make sure everything is crystal clear. This journey will not only help you solve this particular equation but also equip you with the skills to tackle similar problems with confidence. So, buckle up, grab your favorite snack, and let's get started on this exciting mathematical adventure! The initial setup might seem a bit tricky, but trust me, with a systematic approach, we'll crack this code in no time. We will start by isolating the terms involving x and then simplify the equation to something we can easily handle. The beauty of mathematics lies in its logical structure, so let’s leverage that to our advantage. The goal is always to transform the complex into the simple, the unknown into the known. Are you ready to dive in and unleash your inner math wizard? Because I sure am! This process is all about transforming a complex equation into a much simpler form that allows us to find the roots, or solutions, without breaking a sweat. So, let’s begin our adventure of solving quadratic equations and uncovering the secrets hidden within this seemingly complex expression. By the end, you'll be able to conquer similar equations with ease and confidence.
Isolating and Simplifying the Equation
Alright, guys, let's get our hands dirty and start solving this beast of an equation: $\fracx2}{\sqrt{x2-c2}}=\frac{c2}{\sqrt{x2-c2}}+39$. The first step is all about making things look a little less messy. We'll start by getting all the terms with the square root on one side of the equation. To do this, let's subtract $\frac{c2}{\sqrt{x2-c^2}}$ from both sides of the equation. This gives us\sqrt{x2-c2}} - \frac{c2}{\sqrt{x2-c^2}} = 39$. Now, we can simplify the left side since the terms have a common denominator. Combining the fractions, we get $\frac{x^2 - c2}{\sqrt{x2-c^2}} = 39$. See? Already, it's looking a bit cleaner! The next move is to make the square root disappear. Multiply both sides by $\sqrt{x2-c2}$ to eliminate the denominator$. At this point, it’s all about the transformation; the goal is to make it manageable, step by step. This method provides a clear path forward, and the logic is straightforward. Now, we're not just moving terms around; we're strategically simplifying the equation. It's like a puzzle, and each move brings us closer to the solution. The core idea is to isolate the square root and get rid of it. Remember to always keep the equation balanced by performing the same operations on both sides. This ensures that the equality remains true throughout our manipulations. Each step we take brings us one step closer to unveiling the solution and mastering the art of solving quadratic equations. This systematic approach is the secret to conquering even the trickiest equations.
Squaring Both Sides and Solving for x
Okay, team, we've simplified our equation to $x^2 - c^2 = 39\sqrt{x2-c2}$. Now, we want to get rid of that pesky square root once and for all. The best way to do this? Square both sides of the equation. This gives us $(x^2 - c2)2 = (39\sqrt{x2-c2})^2$. Let's expand both sides. On the left side, we have $(x^2 - c2)2 = x^4 - 2c2x2 + c^4$. On the right side, we get $(39\sqrt{x2-c2})^2 = 392(x2 - c^2) = 1521(x^2 - c^2)$. So, our equation now looks like this: $x^4 - 2c2x2 + c^4 = 1521x^2 - 1521c^2$. Now, let's bring everything to one side to get a standard quadratic form (sort of!). Subtract $1521x^2$ and add $1521c^2$ to both sides: $x^4 - 2c2x2 - 1521x^2 + c^4 + 1521c^2 = 0$. Combine like terms: $x^4 - (2c^2 + 1521)x^2 + (c^4 + 1521c^2) = 0$. This might look a bit intimidating, but remember, we're looking for solutions for x. This form, although complex, is still something we can work with. The next part will depend on how we approach it. We can treat this as a quadratic equation in terms of $x^2$. This means we'll try to solve for $x^2$, and then we can find x itself. This is a common trick in algebra; turning a problem into a more familiar one. This step might require a bit of thinking, but it's a critical step in reaching the final solution. The key is to notice that we have terms with $x^4$, which is the square of $x^2$, so we can use substitution. The purpose here is to manipulate the equation to identify the possible values of x. The expansion and simplification are crucial to finding the solutions. We're getting closer to solving this equation and revealing the value of x. Keep up the great work!
Identifying Potential Solutions and Checking the Answers
Alright, let's take a closer look at our equation, $x^4 - (2c^2 + 1521)x^2 + (c^4 + 1521c^2) = 0$. To solve this equation, let's make a substitution. Let $y = x^2$. Then, our equation becomes $y^2 - (2c^2 + 1521)y + (c^4 + 1521c^2) = 0$. Now, we can use the quadratic formula to solve for y: $y = \frac-b \pm \sqrt{b^2 - 4ac}}{2a}$. In our case, $a = 1$, $b = -(2c^2 + 1521)$, and $c = c^4 + 1521c^2$. So, we have}2}$. Let's simplify the discriminant (the part inside the square root)}2}$. Since $\sqrt{2313441} = 1521$, we get2}$. This gives us two possible values for y2} = \frac{2c^2 + 3042}{2} = c^2 + 1521$ and $y_2 = \frac{2c^2 + 1521 - 1521}{2} = \frac{2c^2}{2} = c^2$. Remember, $y = x^2$, so we can find x by taking the square root of y. For $y_1$, $x^2 = c^2 + 1521$, which means $x = \pm \sqrt{c^2 + 1521}$. For $y_2$, $x^2 = c^2$, which means $x = \pm c$. Now, we have four potential solutions, x = -\sqrt{c^2 + 1521}$. However, we must check these solutions in the original equation to ensure they are valid. Let’s carefully examine each potential solution to verify its correctness. It is really important to check these solutions back in the original equation. Each step we've taken so far has brought us closer to unveiling the solutions, and now it's time to test them and see what holds. Remember to be cautious when dealing with square roots, as they can introduce extraneous solutions. This final step is crucial to ensure that we find the accurate solution.
Checking the Solutions
Now, let's plug our potential solutions back into the original equation, $\frac{x2}{\sqrt{x2-c2}}=\frac{c2}{\sqrt{x2-c2}}+39$, to see which ones are valid. First, consider $x = c$. Plugging this in, we get $\frac{c2}{\sqrt{c2-c^2}} = \frac{c^2}{\sqrt{0}}$ which is undefined because we can’t divide by zero. So, $x = c$ is not a solution. Similarly, if we plug in $x = -c$, we get the same result, so $x = -c$ is also not a solution. Now, let's try $x = \sqrt{c^2 + 1521}$. Plugging this into the equation, we get $\frac{(\sqrt{c^2 + 1521})2}{\sqrt{(\sqrt{c2 + 1521})2-c2}} = \frac{c^2 + 1521}{\sqrt{c^2 + 1521 - c^2}} = \frac{c^2 + 1521}{\sqrt{1521}} = \frac{c^2 + 1521}{39}$. Also, $\frac{c2}{\sqrt{(\sqrt{c2 + 1521})2-c2}} + 39 = \frac{c^2}{\sqrt{1521}} + 39 = \frac{c^2}{39} + 39$. For these two expressions to be equal, we would need $\frac{c^2 + 1521}{39} = \frac{c^2}{39} + 39$, which simplifies to $\frac{c^2}{39} + 39 = \frac{c^2}{39} + 39$. This is indeed true. Therefore, $x = \sqrt{c^2 + 1521}$ is a solution. Finally, let's check $x = -\sqrt{c^2 + 1521}$. Because the equation involves $x^2$, this solution will also work. So, this gives us the same result. So, the only valid solution is $x = \pm \sqrt{c^2 + 1521}$. In conclusion, when we have these potential solutions, we must always check back with the original equation. It's an important step. This is a common practice in mathematics; the act of checking our work. We've gone through the process of verifying all possible solutions, and we have successfully found the correct ones. The process ensures that we identify all valid roots and avoid extraneous solutions. This step is a cornerstone of the problem-solving. This process helps us build trust in our findings and shows our work to be accurate.