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The given equation is:

\[
\sqrt{7x - 1} = \sqrt{8x - 9}
\]

To solve this equation, we can eliminate the square roots by squaring both sides:

\[
(\sqrt{7x - 1})^2 = (\sqrt{8x - 9})^2
\]

This simplifies to:

\[
7x - 1 = 8x - 9
\]

Now, solve for \(x\):

1. Subtract \(7x\) from both sides:
   \[
   -1 = x - 9
   \]

2. Add 9 to both sides:
   \[
   8 = x
   \]

So, the solution is \(x = 8\).

Finally, let's verify by substituting \(x = 8\) back into the original equation:

\[
\sqrt{7(8) - 1} = \sqrt{8(8) - 9}
\]

Simplifying both sides:

\[
\sqrt{56 - 1} = \sqrt{64 - 9}
\]

\[
\sqrt{55} = \sqrt{55}
\]

Both sides are equal, so \(x = 8\) is the correct solution.

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How do you handle square roots in more complex equations?
2. What are the steps to solve radical equations that have terms outside the square roots?
3. What happens if squaring introduces extraneous solutions? How do we check for them?
4. How do you solve radical equations with higher-order roots, like cube roots?
5. How can inequalities be handled when square roots are involved?

**Tip:** Always check your solutions when dealing with square root equations, as squaring both sides can introduce extraneous solutions.

Learn how to solve the radical equation √(7x - 1) = √(8x - 9) by squaring both sides to remove the square roots and solving for x. The problem involves key algebraic concepts, such as properties of square roots and linear equations. Suitable for students in Grades 9-10.

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