To solve the quadratic equation [tex]\( x^2 + 2x = 1 \)[/tex] by completing the square, we will follow these steps:
1. Move the constant term to the left side:
[tex]\[
x^2 + 2x - 1 = 0
\][/tex]
2. Complete the square:
For the quadratic term [tex]\( x^2 + 2x \)[/tex], we complete the square by finding a value that makes it a perfect square trinomial. Here, we focus on the term [tex]\( x^2 + 2x \)[/tex].
The formula to complete the square for [tex]\( ax^2 + bx \)[/tex] is [tex]\( x^2 + bx = (x + \frac{b}{2})^2 - (\frac{b}{2})^2 \)[/tex]. For our equation [tex]\( x^2 + 2x \)[/tex]:
- The coefficient [tex]\( b \)[/tex] is 2.
- Calculate [tex]\( \left( \frac{2}{2} \right)^2 = 1 \)[/tex].
Thus,
[tex]\[
x^2 + 2x = (x + 1)^2 - 1
\][/tex]
3. Rewrite the equation:
Substitute this back into the equation [tex]\( x^2 + 2x - 1 = 0 \)[/tex]:
[tex]\[
(x + 1)^2 - 1 - 1 = 0
\][/tex]
Simplify:
[tex]\[
(x + 1)^2 - 2 = 0
\][/tex]
4. Isolate the perfect square:
[tex]\[
(x + 1)^2 = 2
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
Take the square root of both sides:
[tex]\[
x + 1 = \pm \sqrt{2}
\][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[
x = -1 \pm \sqrt{2}
\][/tex]
Hence, we have two solutions:
[tex]\[
x = -1 + \sqrt{2} \quad \text{and} \quad x = -1 - \sqrt{2}
\][/tex]
6. Match these solutions with the given choices:
- [tex]\( x = \sqrt{2} - 1 \)[/tex]
- [tex]\( x = -\sqrt{2} - 1 \)[/tex]
- [tex]\( x = \sqrt{2} + 1 \)[/tex]
- [tex]\( x = -\sqrt{2} + 1 \)[/tex]
- [tex]\( x = 2 \)[/tex]
- [tex]\( x = 0 \)[/tex]
The solutions we found [tex]\( x = -1 + \sqrt{2} \)[/tex] and [tex]\( x = -1 - \sqrt{2} \)[/tex] match with:
[tex]\[
x = \sqrt{2} - 1 \quad \text{(same as \( -1 + \sqrt{2} \))}
\][/tex]
[tex]\[
x = -\sqrt{2} - 1 \quad \text{(same as \( -1 - \sqrt{2} \))}
\][/tex]
Therefore, the correct answers are:
[tex]\[
x = \sqrt{2} - 1 \quad \text{and} \quad x = -\sqrt{2} - 1
\][/tex]
