given that p = 3-2$$ \sqrt2 $$and q = 2-2$$ \sqrt2 $$, find the value of (p+q) / (p-q)

We are given the following expressions for $ p $ and $ q $:

\[
p = 3 - 2\sqrt{2}
\]
\[
q = 2 - 2\sqrt{2}
\]

We need to find the value of $ \frac{p+q}{p-q} $.

### Step 1: Compute $ p + q $
\[
p + q = (3 - 2\sqrt{2}) + (2 - 2\sqrt{2})
\]
Combine like terms:
\[
p + q = (3 + 2) - 2\sqrt{2} - 2\sqrt{2}
\]
\[
p + q = 5 - 4\sqrt{2}
\]

### Step 2: Compute $ p - q $
\[
p - q = (3 - 2\sqrt{2}) - (2 - 2\sqrt{2})
\]
Distribute the negative sign and combine like terms:
\[
p - q = 3 - 2\sqrt{2} - 2 + 2\sqrt{2}
\]
\[
p - q = (3 - 2) + (-2\sqrt{2} + 2\sqrt{2})
\]
\[
p - q = 1
\]

### Step 3: Compute $ \frac{p+q}{p-q} $
We now substitute the values of $ p + q $ and $ p - q $:
\[
\frac{p+q}{p-q} = \frac{5 - 4\sqrt{2}}{1}
\]
\[
\frac{p+q}{p-q} = 5 - 4\sqrt{2}
\]

Thus, the value of $ \frac{p+q}{p-q} $ is:

\[
\boxed{5 - 4\sqrt{2}}
\]

Do you want further details or have any questions? 

Here are some relative questions:
1. What would happen if the radicals were cubed instead of square roots?
2. How does the subtraction of radicals influence the outcome?
3. How do we handle expressions with complex numbers instead of square roots?
4. What if $ p $ and $ q $ were fractions instead of integers plus radicals?
5. How would the solution change if $ p $ and $ q $ had different coefficients for $ \sqrt{2} $?

**Tip**: Simplify expressions with like terms before applying any operations to make the process easier.

Given that p = 3 - 2√2 and q = 2 - 2√2, find the value of (p + q) / (p - q).

Learn how to simplify the expression (p + q) / (p - q) where p and q contain square roots. This step-by-step solution involves algebraic simplification and arithmetic with radicals. Suitable for high school students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation