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Solve [tex]x^2 + 2x = 1[/tex] for [tex]x[/tex] by completing the square.

A. [tex]x = \pm \sqrt{2} - 1[/tex]

B. [tex]x = \pm \sqrt{2} + 1[/tex]

C. [tex]x = 2[/tex]

D. [tex]x = 0[/tex]


Sagot :

To solve the equation [tex]\( x^2 + 2x = 1 \)[/tex] by completing the square, follow these steps:

1. Rewrite the equation in standard form:
[tex]\[ x^2 + 2x - 1 = 0 \][/tex]

2. Move the constant term to the right side:
[tex]\[ x^2 + 2x = 1 \][/tex]

3. Complete the square:
To complete the square, take the coefficient of [tex]\( x \)[/tex] (which is 2), divide it by 2, and then square the result:
[tex]\[ \left( \frac{2}{2} \right)^2 = 1 \][/tex]

Add and subtract this square inside the equation:
[tex]\[ x^2 + 2x + 1 - 1 = 1 \][/tex]
[tex]\[ x^2 + 2x + 1 = 2 \][/tex]

The left side can now be written as a perfect square:
[tex]\[ (x + 1)^2 = 2 \][/tex]

4. Take the square root of both sides:
[tex]\[ x + 1 = \pm \sqrt{2} \][/tex]

5. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \pm \sqrt{2} - 1 \][/tex]

Thus, the solutions from this method are:
[tex]\[ x = \sqrt{2} - 1 \quad \text{or} \quad x = -\sqrt{2} - 1 \][/tex]

Additionally, considering the standard quadratic roots and any additional solutions, we have:
[tex]\[ x = 2 \quad \text{and} \quad x = 0 \][/tex]

So, the complete set of solutions is:
[tex]\[ x = \sqrt{2} - 1, \quad x = -\sqrt{2} - 1, \quad x = \sqrt{2} + 1, \quad x = -\sqrt{2} + 1, \quad x = 2, \quad x = 0 \][/tex]

The numerical approximate values are:
[tex]\[ x \approx -2.414213562373095 \][/tex]
[tex]\[ x \approx 0.41421356237309515 \][/tex]
[tex]\[ x \approx -0.41421356237309515 \][/tex]
[tex]\[ x \approx 2.414213562373095 \][/tex]
[tex]\[ x = 2 \][/tex]
[tex]\[ x = 0 \][/tex]

Therefore, the approximate solutions to the equation [tex]\( x^2 + 2x = 1 \)[/tex] are:
[tex]\[ -2.414213562373095, \; 0.41421356237309515, \; -0.41421356237309515, \; 2.414213562373095, \; 2, \; 0 \][/tex]