By how much does `sqrt(12) + sqrt(18)` exceed `sqrt(3) + sqrt(2)` ?

Answer

To solve the problem of how much `sqrt(12) + sqrt(18)` exceeds `sqrt(3) + sqrt(2)`, we will follow these steps:

### Step 1: Simplify `sqrt(12)` and `sqrt(18)`
- `sqrt(12)` can be simplified as follows:
\[ \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \]
- `sqrt(18)` can be simplified as follows:
\[ \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2} \]

### Step 2: Combine the simplified forms
Now we can combine the simplified forms of `sqrt(12)` and `sqrt(18)`:
\[ \sqrt{12} + \sqrt{18} = 2\sqrt{3} + 3\sqrt{2} \]

### Step 3: Write down `sqrt(3) + sqrt(2)`
The expression `sqrt(3) + sqrt(2)` remains as it is:
\[ \sqrt{3} + \sqrt{2} \]

### Step 4: Subtract `sqrt(3) + sqrt(2)` from `sqrt(12) + sqrt(18)`
Now we will subtract the second expression from the first:
\[ (2\sqrt{3} + 3\sqrt{2}) - (\sqrt{3} + \sqrt{2}) \]

### Step 5: Combine like terms
Now, we will combine the like terms:
- For `sqrt(3)` terms:
\[ 2\sqrt{3} - \sqrt{3} = (2 - 1)\sqrt{3} = 1\sqrt{3} = \sqrt{3} \]
- For `sqrt(2)` terms:
\[ 3\sqrt{2} - \sqrt{2} = (3 - 1)\sqrt{2} = 2\sqrt{2} \]

### Step 6: Write the final result
Combining the results from the previous step, we get:
\[ \sqrt{3} + 2\sqrt{2} \]

Thus, `sqrt(12) + sqrt(18)` exceeds `sqrt(3) + sqrt(2)` by:
\[ \sqrt{3} + 2\sqrt{2} \]