Solving Radical Equations: Finding The Value Of 'u'

Nov 10, 2025 by Admin 52 views

Hey math enthusiasts! Today, we're diving into the world of radical equations to solve for 'u' in the equation: $\sqrt{4 u+12}=u$ . This might seem a bit intimidating at first, but trust me, we'll break it down step by step to make it super clear and easy to understand. So, grab your pencils and let's get started! Our goal here is to isolate u, a real number, and find its value(s) that satisfy the given equation. It's like a treasure hunt, and we're looking for the hidden 'u'. Remember, the key to solving radical equations is to get rid of the square root. How do we do that? By squaring both sides of the equation. This is a fundamental concept in algebra, and understanding it will open doors to solving many other types of equations. We'll also need to be mindful of potential extraneous solutions. These are values that appear to be solutions when we work through the algebra, but they don't actually satisfy the original equation. We'll check our answers at the end to make sure everything is on the up-and-up.

First things first: We have the equation $\sqrt{4 u+12}=u$ . To begin, we want to eliminate that pesky square root sign. The most direct way to do that is to square both sides of the equation. This is a valid algebraic manipulation because as long as we perform the same operation on both sides, the equation remains balanced. By squaring both sides, we're essentially undoing the square root operation. So, let's square both sides: $(\sqrt{4 u+12})^2 = u^2$ . This simplifies to $4u + 12 = u^2$ . Now, this is a quadratic equation! This means that it has the form $au^2 + bu + c = 0$ , where a, b, and c are constants. To solve a quadratic equation, we typically rearrange it so that one side is equal to zero, and then we can either factor it, complete the square, or use the quadratic formula. Let's rearrange our equation to $u^2 - 4u - 12 = 0$ . Now we're cooking with gas, guys! We have a quadratic equation ready to be solved. Let's see if we can factor this bad boy. Factoring is a handy technique that can often simplify the process. We need to find two numbers that multiply to give us -12 (the constant term) and add up to -4 (the coefficient of the u term). After a bit of mental gymnastics (or trial and error), we find that -6 and 2 fit the bill. So, we can factor the quadratic equation as $(u - 6)(u + 2) = 0$ . This factored form is super useful because the product of two factors is zero only if one or both of the factors are zero. This leads us to our next step. Now, we have two possible solutions. Either $u - 6 = 0$ or $u + 2 = 0$ . Solving these gives us $u = 6$ or $u = -2$ . But hold your horses! Before we declare victory and celebrate with ice cream, we need to check these solutions in the original equation to make sure they're valid.

Checking the Solutions: Are They Extraneous?

Alright, folks, now comes the crucial part: checking our solutions. Remember those potential extraneous solutions I mentioned earlier? This is where we find out if any of our solutions are imposters. We have two possible solutions, $u = 6$ and $u = -2$ . We need to plug each of these back into the original equation, $\sqrt{4 u+12}=u$ , to see if they hold true. Let's start with $u = 6$ . Substitute 6 into the equation: $\sqrt{4(6) + 12} = 6$ . Simplifying inside the square root: $\sqrt{24 + 12} = 6$ , which gives us $\sqrt{36} = 6$ . And since the square root of 36 is indeed 6, this solution checks out! So, $u = 6$ is a valid solution. Awesome!

Now, let's try $u = -2$ . Substitute -2 into the original equation: $\sqrt{4(-2) + 12} = -2$ . Simplifying: $\sqrt{-8 + 12} = -2$ , which gives us $\sqrt{4} = -2$ . The square root of 4 is 2, so we have $2 = -2$ . Uh oh... this is not true! This means that $u = -2$ is an extraneous solution. It appeared to be a solution when we were solving the quadratic equation, but it doesn't actually work in the original equation. Extraneous solutions often pop up when we square both sides of an equation because squaring can sometimes introduce values that weren't originally part of the solution set. Therefore, we must discard $u = -2$ as a valid solution. So, what's our final answer? We only have one valid solution: $u = 6$ . Always remember to check your solutions in the original equation, especially when dealing with radical equations. This step is super important to avoid getting tricked by extraneous solutions! It's like double-checking your work to make sure you didn't miss anything. By carefully checking our solutions, we've ensured that our final answer is accurate and valid. This process reinforces the importance of algebraic manipulation and the need for careful verification in mathematics. Great job, everyone! Let's summarize the key steps we took to solve for u and how to avoid making costly mistakes.

Summary of Steps and Key Takeaways

To recap, here's a quick rundown of the steps we followed to solve for u:

Isolate the Radical: Our equation was already set up with the radical term isolated.
Square Both Sides: We squared both sides of the equation to eliminate the square root.
Solve the Resulting Equation: This gave us a quadratic equation, which we factored and solved for two possible values of u.
Check for Extraneous Solutions: We substituted both values back into the original equation to see if they were valid solutions. This is the most crucial step!

Key Takeaways:

When dealing with radical equations, always square both sides to eliminate the radical (square root).
Remember that squaring both sides can introduce extraneous solutions.
Always check your solutions in the original equation to verify their validity.
Factoring, as demonstrated, simplifies the quadratic equation and leads to solutions.

By following these steps and remembering these key takeaways, you'll be well-equipped to tackle any radical equation that comes your way. Keep practicing, and you'll become a pro at solving for u in no time. If you have any questions or want to try another problem, feel free to ask. Keep up the great work, and keep exploring the wonderful world of mathematics!

Conclusion

So, there you have it, folks! We've successfully navigated the challenges of solving a radical equation and emerged victorious, finding that the real solution for u is 6. Remember the importance of checking your answers! Keep practicing these concepts, and you will become more comfortable with radical equations. Don't be afraid to experiment, and you'll find that math can actually be pretty fun. The key is breaking down complex problems into smaller, manageable steps, and always double-checking your work. Happy solving, and keep those math muscles flexed!