If x =$\frac{1}{1+\sqrt{2}}$, then the value of $x^2 + 2x + 3$ is :

Q: If x =$\frac{1}{1+\sqrt{2}}$, then the value of $x^2 + 2x + 3$ is :

To determine the value of $ x^2 + 2x + 3 $ given that $ x = \frac{1}{1+\sqrt{2}} $, we will first simplify the expression for $ x $. ### Step 1: Simplify $ x $ Given: \[ x = \frac{1}{1+\sqrt{2}} \] To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator: \[ x = \frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} \] The denominator simplifies as follows: \[ (1+\sqrt{2})(1-\sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1 \] Thus: \[ x = \frac{1-\sqrt{2}}{-1} = \sqrt{2} - 1 \] ### Step 2: Calculate $ x^2 + 2x + 3 $ Now substitute $ x = \sqrt{2} - 1 $ into the expression $ x^2 + 2x + 3 $: 1. Calculate $ x^2 $: \[ x^2 = (\sqrt{2} - 1)^2 = (\sqrt{2})^2 - 2 \cdot \sqrt{2} \cdot 1 + 1^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} \] 2. Calculate $ 2x $: \[ 2x = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2 \] 3. Substitute into $ x^2 + 2x + 3 $: \[ x^2 + 2x + 3 = (3 - 2\sqrt{2}) + (2\sqrt{2} - 2) + 3 \] Combine the terms: \[ = 3 - 2\sqrt{2} + 2\sqrt{2} - 2 + 3 \] \[ = 3 - 2 + 3 \] \[ = 4 \] Thus, the value of $ x^2 + 2x + 3 $ is 4. ### Conclusion - **Option 3 (4)** is correct because the calculated value of $ x^2 + 2x + 3 $ is indeed 4. - **Option 1 (0)**, **Option 2 (1)**, and **Option 4 (2)** are incorrect because they do not match the calculated value of 4. Therefore, the correct answer is Option 3.

Question

Accepted Answer

To determine the value of $ x^2 + 2x + 3 $ given that $ x = \frac{1}{1+\sqrt{2}} $, we will first simplify the expression for $ x $.

### Step 1: Simplify $ x $

Given:
$$ x = \frac{1}{1+\sqrt{2}} $$

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator:
$$ x = \frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} $$

The denominator simplifies as follows:
$$ (1+\sqrt{2})(1-\sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1 $$

Thus:
$$ x = \frac{1-\sqrt{2}}{-1} = \sqrt{2} - 1 $$

### Step 2: Calculate $ x^2 + 2x + 3 $

Now substitute $ x = \sqrt{2} - 1 $ into the expression $ x^2 + 2x + 3 $:

1. Calculate $ x^2 $:
   $$ x^2 = (\sqrt{2} - 1)^2 = (\sqrt{2})^2 - 2 \cdot \sqrt{2} \cdot 1 + 1^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} $$

2. Calculate $ 2x $:
   $$ 2x = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2 $$

3. Substitute into $ x^2 + 2x + 3 $:
   $$ x^2 + 2x + 3 = (3 - 2\sqrt{2}) + (2\sqrt{2} - 2) + 3 $$

Combine the terms:
$$ = 3 - 2\sqrt{2} + 2\sqrt{2} - 2 + 3 $$
$$ = 3 - 2 + 3 $$
$$ = 4 $$

Thus, the value of $ x^2 + 2x + 3 $ is 4.

### Conclusion

- **Option 3 (4)** is correct because the calculated value of $ x^2 + 2x + 3 $ is indeed 4.
- **Option 1 (0)**, **Option 2 (1)**, and **Option 4 (2)** are incorrect because they do not match the calculated value of 4.

Therefore, the correct answer is Option 3.