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The problem you've uploaded asks to prove the following equation given two non-zero real numbers \( a \) and \( b \):

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

Additionally, you are given that \( \frac{b}{a} = 9 \).

### Step-by-Step Solution:

1. **Given Equation:**

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

2. **Substitute the relation \( \frac{b}{a} = 9 \)**, or equivalently \( b = 9a \):

   \[
   a + 9a + 2\sqrt{a \cdot 9a} = 2(9a - a)
   \]

   Simplify:

   \[
   a + 9a + 2\sqrt{9a^2} = 2(8a)
   \]

3. **Simplify further:**

   The term \( \sqrt{9a^2} = 3a \), so:

   \[
   a + 9a + 2(3a) = 16a
   \]

   Which becomes:

   \[
   a + 9a + 6a = 16a
   \]

   Simplifying the left-hand side:

   \[
   16a = 16a
   \]

This shows that the given equation holds true.

---

Would you like more details or have any questions? Here are five related questions:

1. How can you manipulate square roots in algebraic expressions?
2. Why is the substitution \( \frac{b}{a} = 9 \) crucial to the proof?
3. Can this method be applied to other similar problems involving roots?
4. What happens if \( a \) and \( b \) were negative in this context?
5. How would this change if the equation were cubic instead of quadratic?

**Tip:** Always verify your algebraic manipulations step-by-step to avoid small mistakes, especially when dealing with square roots.

Given two non-zero real numbers a and b such that: a + b + 2√ab = 2(b - a), show that b/a = 9.

This problem involves solving an algebraic equation with radicals, where a + b + 2√ab = 2(b - a), and it is given that b/a = 9. The solution demonstrates key algebraic manipulations and the use of square roots.

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