We are tasked with solving the equation:

\[
1200 = 50 \left( \frac{2}{3}k \right)k
\]

### Step 1: Simplify the equation
First, distribute the \( 50 \) to the terms inside the parentheses:

\[
1200 = 50 \times \frac{2}{3}k \times k
\]
This simplifies to:

\[
1200 = \frac{100}{3} k^2
\]

### Step 2: Eliminate the fraction
Multiply both sides of the equation by 3 to eliminate the denominator:

\[
3600 = 100k^2
\]

### Step 3: Solve for \( k^2 \)
Now, divide both sides by 100:

\[
36 = k^2
\]

### Step 4: Solve for \( k \)
Take the square root of both sides:

\[
k = \sqrt{36}
\]

This gives:

\[
k = 6
\]

Thus, the solution is \( k = 6 \).

Would you like any further details or explanations?

### Related questions:
1. How would the solution change if the equation had been \( 1200 = 50 \left( \frac{3}{2}k \right)k \)?
2. What would happen if \( k \) were negative in this equation?
3. Can you show how to graph \( 50\left( \frac{2}{3}k \right)k \)?
4. How does factoring quadratic expressions relate to solving this problem?
5. How would you solve \( 1200 = 50 \left( \frac{2}{3}k \right)^2 \)?

### Tip:
Always check for extraneous solutions when dealing with square roots or other powers of variables!

Discover how to solve the quadratic equation 1200 = 50(2/3k)k with a clear, step-by-step guide. This solution involves key algebraic techniques such as the distributive property and solving for k using square roots. Ideal for students in Grades 9-10.

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