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The given equation is:

\[
\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{a}{x}
\]

We can solve this equation by manipulating the left-hand side. Let's start by simplifying the numerator and denominator separately.

### Step 1: Multiply by the conjugate
We multiply both the numerator and the denominator by the conjugate of the denominator to eliminate the square roots in the denominator. The conjugate of \(\sqrt{a + x} - \sqrt{a - x}\) is \(\sqrt{a + x} + \sqrt{a - x}\). So:

\[
\frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} - \sqrt{a - x} \right)} \times \frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)}{\left( \sqrt{a + x} + \sqrt{a - x} \right)}
\]

This simplifies to:

\[
\frac{\left( \sqrt{a + x} + \sqrt{a - x} \right)^2}{\left( \sqrt{a + x} \right)^2 - \left( \sqrt{a - x} \right)^2}
\]

### Step 2: Simplify the denominator
Using the difference of squares formula, we simplify the denominator:

\[
\left( \sqrt{a + x} \right)^2 - \left( \sqrt{a - x} \right)^2 = (a + x) - (a - x) = a + x - a + x = 2x
\]

Thus, the denominator becomes \(2x\).

### Step 3: Expand the numerator
Now expand the numerator \(\left( \sqrt{a + x} + \sqrt{a - x} \right)^2\):

\[
\left( \sqrt{a + x} + \sqrt{a - x} \right)^2 = (\sqrt{a + x})^2 + 2\sqrt{(a + x)(a - x)} + (\sqrt{a - x})^2
\]

Simplifying:

\[
= (a + x) + 2\sqrt{(a + x)(a - x)} + (a - x)
\]

The \(x\)'s cancel out, leaving:

\[
a + a + 2\sqrt{(a + x)(a - x)} = 2a + 2\sqrt{a^2 - x^2}
\]

### Step 4: Simplify the full expression
Now, the left-hand side becomes:

\[
\frac{2a + 2\sqrt{a^2 - x^2}}{2x}
\]

We can factor out the 2 in the numerator:

\[
\frac{2(a + \sqrt{a^2 - x^2})}{2x} = \frac{a + \sqrt{a^2 - x^2}}{x}
\]

### Step 5: Compare with the right-hand side
We are given that the right-hand side is \(\frac{a}{x}\). So:

\[
\frac{a + \sqrt{a^2 - x^2}}{x} = \frac{a}{x}
\]

For this to hold true, the term \(\sqrt{a^2 - x^2}\) must be equal to 0, which implies:

\[
a^2 - x^2 = 0 \quad \Rightarrow \quad a^2 = x^2 \quad \Rightarrow \quad a = x
\]

### Conclusion
The solution to the equation is \(a = x\).

---

Would you like more details on any step, or do you have any further questions?

Here are 5 related questions to explore:
1. How do you find the conjugate of a binomial containing square roots?
2. What is the difference of squares formula, and how is it used?
3. How do you simplify expressions involving square roots?
4. Can you solve other equations by multiplying by conjugates?
5. What happens if you have more terms under the square roots?

**Tip:** Always check for common patterns like the difference of squares when simplifying complex square root expressions!

Solve the equation: \(\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \frac{a}{x}\)

Explore how to solve a radical equation involving square roots and algebraic simplifications. Learn how to use the difference of squares formula and multiply by conjugates to simplify expressions and find the solution, \(a = x\). Suitable for students in Grades 10-12.

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