We are given the equation $x = 3 - 2\sqrt{2}$. The goal is to find:

$\frac{\sqrt{x + 1}}{\sqrt{x}}$

Let's proceed step by step:

Step 1: Simplify $x + 1$

We know that $x = 3 - 2\sqrt{2}$. So,

$x + 1 = (3 - 2\sqrt{2}) + 1 = 4 - 2\sqrt{2}$

Step 2: Find the square roots of $x$ and $x + 1$

We need to find the square roots of both $x = 3 - 2\sqrt{2}$ and $x + 1 = 4 - 2\sqrt{2}$.

Square root of $x$:

$x = 3 - 2\sqrt{2} = (\sqrt{2} - 1)^2$ So, $\sqrt{x} = \sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$

Square root of $x + 1$:

$x + 1 = 4 - 2\sqrt{2} = (\sqrt{2} + 1)^2$ So, $\sqrt{x + 1} = \sqrt{4 - 2\sqrt{2}} = \sqrt{(\sqrt{2} + 1)^2} = \sqrt{2} + 1$

Step 3: Calculate $\frac{\sqrt{x + 1}}{\sqrt{x}}$

Now we calculate the expression:

$\frac{\sqrt{x + 1}}{\sqrt{x}} = \frac{\sqrt{2} + 1}{\sqrt{2} - 1}$

To simplify this, multiply the numerator and denominator by $\sqrt{2} + 1$ (the conjugate of the denominator):

$\frac{\sqrt{2} + 1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} + 1)^2}{(\sqrt{2})^2 - (1)^2}$

Simplifying the numerator and denominator:

• Numerator: $(\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$
• Denominator: $(\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$

Thus, we get:

$\frac{3 + 2\sqrt{2}}{1} = 3 + 2\sqrt{2}$

So, the final answer is:

$\boxed{3 + 2\sqrt{2}}$

Would you like more details on any step or have questions? Here are 5 related questions:

1. How do you simplify expressions with square roots in the numerator and denominator?
2. What is the process of rationalizing denominators?
3. How can we simplify expressions involving binomials like $(\sqrt{2} - 1)^2$?
4. Why do we use conjugates when simplifying square root expressions?
5. Can you explain how $\sqrt{3 - 2\sqrt{2}}$ simplifies to $\sqrt{2} - 1$?

Tip: Always check for patterns such as perfect squares when simplifying complex square root expressions.