If the domain of this Sqrt(2x + 3|x+2| + Kx) Is all numbers then solve for K

To solve for \( K \) such that the domain of the function \( f(x) = \sqrt{2x + 3|x + 2| + Kx} \) is all real numbers, we need to ensure that the expression inside the square root is always non-negative. This is because the square root function is only defined for non-negative numbers.

### Step 1: Analyze the expression inside the square root
The expression inside the square root is:
\[
2x + 3|x + 2| + Kx
\]
For this to be defined for all \( x \), it must always be greater than or equal to zero:
\[
2x + 3|x + 2| + Kx \geq 0
\]

### Step 2: Consider two cases for the absolute value

#### Case 1: \( x + 2 \geq 0 \) (i.e., \( x \geq -2 \))
In this case, \( |x + 2| = x + 2 \), and the expression becomes:
\[
2x + 3(x + 2) + Kx = 2x + 3x + 6 + Kx = (5 + K)x + 6
\]
For the domain to include all values of \( x \geq -2 \), we must ensure that:
\[
(5 + K)x + 6 \geq 0 \text{ for all } x \geq -2
\]
Let's check this condition.

- When \( x = -2 \), the expression becomes:
\[
(5 + K)(-2) + 6 = -2(5 + K) + 6 = -10 - 2K + 6 = -4 - 2K
\]
For this to be non-negative:
\[
-4 - 2K \geq 0 \implies -2K \geq 4 \implies K \leq -2
\]

Next, consider large values of \( x \). For the expression to be non-negative as \( x \to \infty \), we require that \( 5 + K \geq 0 \), so:
\[
K \geq -5
\]

#### From this case, we get the condition: \( -5 \leq K \leq -2 \).

#### Case 2: \( x + 2 < 0 \) (i.e., \( x < -2 \))
In this case, \( |x + 2| = -(x + 2) = -x - 2 \), and the expression becomes:
\[
2x + 3(-x - 2) + Kx = 2x - 3x - 6 + Kx = (-1 + K)x - 6
\]
For the domain to include \( x < -2 \), we require that:
\[
(-1 + K)x - 6 \geq 0
\]
Let's check this condition.

- For large negative values of \( x \), the term \( (-1 + K)x \) dominates, so for the expression to be non-negative, we need \( -1 + K \leq 0 \), meaning:
\[
K \leq 1
\]

Thus, the condition for \( x < -2 \) becomes \( K \leq 1 \).

### Step 3: Combine the conditions
From Case 1, we have \( -5 \leq K \leq -2 \), and from Case 2, we have \( K \leq 1 \). The common solution is:
\[
-5 \leq K \leq -2
\]

### Final Answer:
The value of \( K \) must satisfy \( -5 \leq K \leq -2 \) in order for the domain of the function \( \sqrt{2x + 3|x + 2| + Kx} \) to be all real numbers.

Would you like further clarification on any of the steps? 

Here are five related questions:
1. How do you solve inequalities involving absolute values?
2. What happens when a function’s domain is restricted by a square root?
3. How do absolute value functions behave in different intervals?
4. Can \( K \) take any other values outside the range \([-5, -2]\)?
5. How would the graph of this function look like for different values of \( K \)?

**Tip:** When dealing with absolute value functions, it’s important to split the function into cases to handle the sign changes properly.

If the domain of this Sqrt(2x + 3|x+2| + Kx) is all numbers, then solve for K

Explore how to solve for K in the function Sqrt(2x + 3|x + 2| + Kx) such that its domain includes all real numbers. This problem addresses square roots, absolute values, and inequalities, with step-by-step explanations. Suitable for high school students in Grades 10-12.

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