Hey guys! Today, we're diving into a fun little math problem that involves solving an equation with a square root and figuring out if our solution is the real deal or an imposter, also known as an extraneous solution. Extraneous solutions can sometimes pop up when we're dealing with radicals, so it's super important to know how to spot them. Let's get started!
The Equation: √[3x+4] = 5
So, the equation we're tackling is √[3x + 4] = 5. This looks a bit intimidating at first glance, but don't worry, we'll break it down step by step. The main thing to remember when you see a square root is that we need to get rid of it to solve for x. And how do we do that? By squaring both sides of the equation, of course! Think of it like this: the square root and the square are like dance partners – they undo each other's moves. Squaring both sides helps us eliminate the radical and simplifies our equation, making it easier to work with. This method allows us to transform the equation into a more manageable form where we can isolate x and find its value. Remember, whatever operation you perform on one side of the equation, you must perform on the other to maintain the balance and ensure the equation remains true. This fundamental principle of algebraic manipulation is crucial for solving any equation accurately. As we proceed, pay close attention to each step, ensuring you understand the logic behind it. This will not only help you solve this particular equation but also equip you with the skills to tackle similar problems in the future. Mathematics, at its core, is about understanding the underlying principles and applying them consistently. So, let's dive in and see how squaring both sides helps us unravel this equation.
Step 1: Squaring Both Sides
To eliminate the square root, we'll square both sides of the equation. Squaring both sides is a crucial step because it helps us get rid of the radical sign, making the equation much easier to solve. Remember, what you do to one side of the equation, you must do to the other to maintain the balance. This principle is fundamental in algebra and ensures that the equality remains valid throughout the solving process. When we square the left side, the square root and the square cancel each other out, leaving us with just the expression inside the square root. On the right side, we simply square the number, which is a straightforward arithmetic operation. This step transforms the equation from one involving a square root to a simple linear equation, which we can then solve using basic algebraic techniques. It's important to perform this step accurately because any mistake here will propagate through the rest of the solution, potentially leading to an incorrect answer. Once we've squared both sides, we can proceed to isolate the variable x, which is our ultimate goal. So, let's carry out this step carefully and see how it simplifies the equation.
(√[3x + 4])² = 5²
This simplifies to:
3x + 4 = 25
Step 2: Isolating x
Now we've got a nice, simple linear equation. To isolate x, we first need to get rid of the '+ 4' on the left side. We do this by subtracting 4 from both sides. Subtracting 4 from both sides maintains the balance of the equation, ensuring that the equality remains valid. This is a fundamental principle in algebra: whatever operation you perform on one side, you must perform on the other. By subtracting 4, we are essentially undoing the addition, which moves us closer to isolating the term with x. This step is crucial because it simplifies the equation further, making it easier to see the next step required to solve for x. Remember, our goal is to get x by itself on one side of the equation, so we need to systematically eliminate any other terms or coefficients that are attached to it. After performing this subtraction, we will have a simpler equation that involves only the term with x and a constant. This will bring us one step closer to finding the value of x. So, let's perform this subtraction carefully and see how it helps us isolate x.
3x + 4 - 4 = 25 - 4
This gives us:
3x = 21
Next, to get x all by itself, we need to divide both sides by 3. Dividing both sides by 3 is the final step in isolating x. Remember, the coefficient of x is the number that multiplies it, and to get x alone, we need to divide both sides of the equation by this coefficient. Just like with addition and subtraction, division must be performed on both sides of the equation to maintain balance. This ensures that the equality remains true and that we are solving for the correct value of x. By dividing by 3, we are undoing the multiplication, which leaves x isolated on one side of the equation. Once we perform this division, we will have the value of x, which is the solution to our equation. However, we're not quite done yet! We still need to check this solution to make sure it's not extraneous. So, let's complete this division and then move on to the crucial step of checking our answer.
3x / 3 = 21 / 3
So,
x = 7
Step 3: Checking for Extraneous Solutions
Alright, we've found a possible solution: x = 7. But hold your horses! We're not done yet. When dealing with square roots, we always need to check our solution(s) to make sure they actually work in the original equation. Sometimes, we might find a solution that seems right but doesn't fit when we plug it back in. These sneaky solutions are called extraneous solutions, and they're like the imposters of the math world. Checking for extraneous solutions is a crucial step when dealing with equations that involve radicals or rational expressions. This is because the operations we perform to solve these equations, such as squaring both sides, can sometimes introduce solutions that do not satisfy the original equation. These extraneous solutions are not valid solutions to the problem, and it's important to identify and discard them. The process of checking involves substituting the potential solution back into the original equation and verifying whether it makes the equation true. If the equation holds true, then the solution is valid. However, if the equation does not hold true, then the solution is extraneous and must be rejected. This step ensures that we only accept solutions that are genuine and satisfy the conditions of the original problem. So, let's take a moment to check our solution x = 7 to make sure it's the real deal and not an extraneous one.
Let's plug x = 7 back into the original equation:
√[3(7) + 4] = 5
√[21 + 4] = 5
√[25] = 5
5 = 5
Woohoo! It checks out. This means that x = 7 is a valid solution, not an extraneous one. Plugging the value back into the original equation is a critical step to ensure that our solution makes sense within the context of the problem. When we substitute x = 7 into the original equation √[3x + 4] = 5, we replace x with 7 and simplify the expression. We first multiply 3 by 7, which gives us 21. Then, we add 4 to 21, resulting in 25. The square root of 25 is 5, which matches the right side of the equation. This confirms that x = 7 is a valid solution because it satisfies the original equation. If, after substituting and simplifying, the left side of the equation did not equal the right side, then x = 7 would have been an extraneous solution, and we would have had to discard it. However, since both sides of the equation are equal, we can confidently say that x = 7 is the correct solution to the equation. This process of verifying solutions is a fundamental aspect of problem-solving in mathematics and helps us avoid errors and ensure accuracy.
Conclusion
So, we solved the equation √[3x + 4] = 5 and found that x = 7. We also checked our solution and confirmed that it is not extraneous. High five! Solving equations with radicals might seem tricky at first, but with a little practice, you'll be a pro in no time. Remember to always check for extraneous solutions to make sure your answer is the real deal. You've got this! Solving this equation involved a few key steps, each building upon the previous one. First, we squared both sides of the equation to eliminate the square root, which transformed the equation into a more manageable form. Then, we isolated x by performing algebraic operations, such as subtracting and dividing, ensuring that we maintained the balance of the equation. Finally, we checked our solution by substituting it back into the original equation to verify that it was not extraneous. This process not only helped us find the solution but also reinforced the importance of checking our work, especially when dealing with radicals. Extraneous solutions can arise due to the nature of squaring both sides of an equation, which can sometimes introduce solutions that do not satisfy the original equation. By checking our answer, we can be confident that we have found the correct solution to the problem. Mathematics is a journey of discovery, and each problem we solve adds to our understanding and skills. So, keep practicing and exploring, and you'll become more proficient in solving even the most challenging equations. Well done on working through this problem with me! Let’s keep exploring the exciting world of mathematics together.