https://file.aiquickdraw.com/mathbot/file/17298262256202i89k0y6.jpg

Let's break down the equation and solve for \( d \).

The equation provided is:

\[
5.972 \times \frac{10^{24}}{7.34 \times 10^{22}} = \left( \frac{d}{3.82 \times 10^8} - d \right)^2
\]

### Step 1: Simplify the left-hand side

We can simplify the left-hand side of the equation:

\[
5.972 \times \frac{10^{24}}{7.34 \times 10^{22}} = 5.972 \times \frac{10^{24}}{7.34 \times 10^{22}} = 5.972 \times 1.3629 = 8.1356
\]

So the equation becomes:

\[
8.1356 = \left( \frac{d}{3.82 \times 10^8} - d \right)^2
\]

### Step 2: Take the square root of both sides

Take the square root of both sides to eliminate the square on the right-hand side:

\[
\sqrt{8.1356} = \frac{d}{3.82 \times 10^8} - d
\]

\[
2.8522 = \frac{d}{3.82 \times 10^8} - d
\]

### Step 3: Solve for \( d \)

We now have a linear equation:

\[
2.8522 = \frac{d}{3.82 \times 10^8} - d
\]

Multiply through by \( 3.82 \times 10^8 \) to get rid of the denominator:

\[
2.8522 \times 3.82 \times 10^8 = d - 3.82 \times 10^8 \times d
\]

Simplifying:

\[
1.089 \times 10^9 = d - 3.82 \times 10^8 d
\]

Factor out \( d \):

\[
1.089 \times 10^9 = d (1 - 3.82 \times 10^8)
\]

Solve for \( d \):

\[
d = \frac{1.089 \times 10^9}{1 - 3.82 \times 10^8}
\]

This will give the solution for \( d \). 

Would you like me to compute the exact value numerically? Let me know if you need more details! 

Here are 5 questions based on this:

1. How does taking the square root help in solving quadratic equations?
2. Can we approximate the behavior of the function if \( d \) is very large or very small?
3. What real-world scenario might this equation represent?
4. Could this equation represent the balance between two forces? What might they be?
5. How would we handle this equation if there were additional terms involving \( d^2 \)?

**Tip**: Always check if your variable appears linearly or quadratically when simplifying. This helps in deciding whether to factor or take roots.

Learn how to solve a quadratic equation in the context of a gravitational equation. This problem demonstrates key algebraic techniques like simplifying ratios, applying square roots, and solving quadratic equations. Suitable for high school physics and mathematics students.

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