Solving The Equation: $3=\left|\frac{4-3 K+3}{\sqrt{4+k^2}}\right|$

$$

Hey guys! Let's dive into solving this interesting mathematical equation. It looks a bit intimidating at first, but we'll break it down step by step. Our main goal here is to find the value(s) of k that satisfy the equation. We'll use a combination of algebraic manipulation and a bit of logical thinking to get there. So, buckle up and let's get started!

Understanding the Equation

At the heart of this problem is the equation $3=\left|\frac{4-3 k+3}{\sqrt{4+k^2}}\right|$. To truly solve it, we need to first understand what every part represents. The left side is straightforward – it's just the number 3. But the right side is more complex, involving an absolute value, a fraction, and a square root. Let's dissect each component:

• Absolute Value: The vertical bars around the fraction, denoted as $|...|$, signify the absolute value. This means that the expression inside the bars will always be positive, regardless of whether the fraction itself is positive or negative. For example, $|3| = 3$ and $|-3| = 3$. In our context, this implies that we need to consider both positive and negative scenarios of the fraction.
• Fraction: We have a fraction with a numerator of $4 - 3k + 3$ and a denominator of $\sqrt{4 + k^2}$. Simplifying the numerator, we get $7 - 3k$. The denominator involves a square root, which introduces some constraints. Since the square root of a number is always non-negative, the denominator will always be positive. This is crucial because it simplifies our analysis of the sign of the fraction.
• Square Root: The term $\sqrt{4 + k^2}$ contains a square root. We know that the expression inside a square root must be non-negative. However, in this case, $4 + k^2$ will always be positive for any real value of k, because $k^2$ is always non-negative, and adding 4 ensures it's strictly positive. This is a relief because we don't have to worry about restricting the possible values of k based on the square root.

So, in essence, we're dealing with a situation where the absolute value of a fraction involving k is equal to 3. This means the fraction itself can be either 3 or -3. This is the key insight that will guide our next steps.

Breaking Down the Absolute Value

The absolute value in the equation $3=\left|\frac{7-3 k}{\sqrt{4+k^2}}\right|$ introduces a critical branching point in our solution. Remember, the absolute value of a number is its distance from zero, which means it can be either the number itself (if it's positive) or the negation of the number (if it's negative). Therefore, to solve the equation, we need to consider two separate cases:

Case 1: The expression inside the absolute value is positive or zero.

In this case, the absolute value simply removes the bars, and we are left with the equation:

$\frac{7-3 k}{\sqrt{4+k^2}} = 3$

This equation states that the fraction is equal to 3. This is a direct translation of the absolute value, assuming the expression inside is non-negative. To solve this, we'll need to get rid of the fraction by multiplying both sides by the denominator. This will lead us to a more manageable equation that we can solve for k.

Case 2: The expression inside the absolute value is negative.

In this scenario, the absolute value changes the sign of the expression. So, we have:

$-\frac{7-3 k}{\sqrt{4+k^2}} = 3$

This is equivalent to:

$\frac{7-3 k}{\sqrt{4+k^2}} = -3$

This equation tells us that the fraction is equal to -3. It's crucial to consider this case because the absolute value makes both positive and negative results of the fraction valid solutions. Again, we'll tackle this by multiplying both sides by the denominator and then solving for k.

By addressing these two cases separately, we ensure that we capture all possible solutions for k. Each case will lead us to a different algebraic equation, which we'll need to solve independently. The solutions from both cases will then be combined to give us the complete solution set for the original equation. This careful approach, dictated by the absolute value, is fundamental to solving equations of this type.

Solving Case 1: $\frac{7-3 k}{\sqrt{4+k^2}} = 3$

Alright, let's tackle the first case. We've got the equation $\frac{7-3 k}{\sqrt{4+k^2}} = 3$. The first order of business is to get rid of that pesky square root in the denominator. To do this, we'll multiply both sides of the equation by $\sqrt{4+k^2}$:

$7 - 3k = 3\sqrt{4 + k^2}$

Now, to eliminate the square root entirely, we'll square both sides of the equation. Remember, squaring both sides can sometimes introduce extraneous solutions, so we'll need to check our answers later to make sure they actually work in the original equation:

$(7 - 3k)^2 = (3\sqrt{4 + k^2})^2$

Expanding both sides, we get:

$49 - 42k + 9k^2 = 9(4 + k^2)$

$49 - 42k + 9k^2 = 36 + 9k^2$

Notice that the $9k^2$ terms cancel out on both sides, which simplifies things nicely. We're left with:

$49 - 42k = 36$

Now, let's isolate k. Subtract 49 from both sides:

$-42k = -13$

Finally, divide both sides by -42:

$k = \frac{13}{42}$

So, we have a potential solution: $k = \frac{13}{42}$. But remember, we squared both sides, so we need to check if this solution is valid. We'll plug it back into the original equation $\frac{7-3 k}{\sqrt{4+k^2}} = 3$ to verify. This is a crucial step to ensure we haven't introduced any extraneous solutions.

Solving Case 2: $\frac{7-3 k}{\sqrt{4+k^2}} = -3$

Now, let's move on to the second case, where $\frac{7-3 k}{\sqrt{4+k^2}} = -3$. Just like in the first case, we want to get rid of the square root. We'll start by multiplying both sides of the equation by $\sqrt{4+k^2}$:

$7 - 3k = -3\sqrt{4 + k^2}$

To eliminate the square root, we'll square both sides of the equation. Again, we need to keep in mind that squaring both sides can introduce extraneous solutions, so we'll need to check our answers later:

$(7 - 3k)^2 = (-3\sqrt{4 + k^2})^2$

Expanding both sides, we get:

$49 - 42k + 9k^2 = 9(4 + k^2)$

$49 - 42k + 9k^2 = 36 + 9k^2$

Notice that this is the exact same equation we obtained in Case 1 after squaring both sides! This means the $9k^2$ terms will cancel out again, leaving us with:

$49 - 42k = 36$

Subtracting 49 from both sides:

$-42k = -13$

Dividing both sides by -42:

$k = \frac{13}{42}$

Wait a minute! We got the same potential solution as in Case 1: $k = \frac{13}{42}$. This is interesting, but it doesn't mean we only have one solution. It just means that the algebraic steps led us to the same value. We still need to check this solution in the original equation for Case 2, which is $\frac{7-3 k}{\sqrt{4+k^2}} = -3$, to see if it holds true. This is a crucial step to confirm whether it's a valid solution for this specific case.

Checking for Extraneous Solutions

Okay, we've found a potential solution, $k = \frac{13}{42}$, but we need to be absolutely sure it works in our original equation, especially because we squared both sides during our solving process. This is the crucial step of checking for extraneous solutions. Let's start by plugging $k = \frac{13}{42}$ back into the original equation:

$3=\left|\frac{7-3 k}{\sqrt{4+k^2}}\right|$

Substitute $k = \frac{13}{42}$:

$3=\left|\frac{7-3(\frac{13}{42})}{\sqrt{4+(\frac{13}{42})^2}}\right|$

Let's simplify the numerator first:

$7 - 3(\frac{13}{42}) = 7 - \frac{13}{14} = \frac{98 - 13}{14} = \frac{85}{14}$

Now, let's simplify the denominator:

$\sqrt{4+(\frac{13}{42})^2} = \sqrt{4 + \frac{169}{1764}} = \sqrt{\frac{7056 + 169}{1764}} = \sqrt{\frac{7225}{1764}} = \frac{85}{42}$

Now, plug these simplified values back into the absolute value expression:

$\left|\frac{\frac{85}{14}}{\frac{85}{42}}\right| = \left|\frac{85}{14} \cdot \frac{42}{85}\right| = |3| = 3$

Great! It works. $k = \frac{13}{42}$ satisfies the original equation. However, we need to remember that we had two cases. We checked this solution against the original absolute value equation, but let's check it specifically against the equations we derived for each case:

Case 1: $\frac{7-3 k}{\sqrt{4+k^2}} = 3$

We already did this check implicitly when we checked the absolute value equation, and it worked out. So, $k = \frac{13}{42}$ is a valid solution for Case 1.

Case 2: $\frac{7-3 k}{\sqrt{4+k^2}} = -3$

Let's plug $k = \frac{13}{42}$ into this equation:

$\frac{7-3(\frac{13}{42})}{\sqrt{4+(\frac{13}{42})^2}} = \frac{\frac{85}{14}}{\frac{85}{42}} = 3$

But we need this to equal -3 for Case 2 to be satisfied! Since it equals 3, $k = \frac{13}{42}$ is not a solution for Case 2.

This is super important! Even though it worked for the absolute value equation, it doesn't work for the specific equation in Case 2. This means it's an extraneous solution for Case 2.

Final Solution

After all the calculations and checks, we've arrived at the final solution. We solved the equation $3=\left|\frac{4-3 k+3}{\sqrt{4+k^2}}\right|$ by breaking it down into two cases based on the absolute value. We found a potential solution, $k = \frac{13}{42}$, but we had to carefully check it for extraneous solutions.

After plugging $k = \frac{13}{42}$ back into the original equation, we confirmed that it is indeed a valid solution. However, when we checked it against the individual cases, we found that it only satisfies Case 1 and not Case 2. This is a crucial observation because it highlights the importance of checking solutions in the context of the specific cases we created to handle the absolute value.

Therefore, the final solution to the equation $3=\left|\frac{4-3 k+3}{\sqrt{4+k^2}}\right|$ is:

$k = \frac{13}{42}$

This whole process demonstrates the importance of careful algebraic manipulation and thorough checking when dealing with absolute values and square roots. We had to be mindful of potential extraneous solutions and ensure that our final answer truly satisfies the original equation.

So, there you have it, guys! We've successfully navigated through this equation, step by step. Remember, math problems like these might seem daunting at first, but with a clear strategy and a bit of perseverance, you can conquer them. Keep practicing, and you'll become a math whiz in no time!