Question

Complete the square to solve the following quadratic equation: x^2+4x=41
.(1 point)
Responses

A) x=−2+5√3, −2−5√3
x equals negative 2 plus 5 Start Root 3 End Root comma negative 2 minus 5 Start Root 3 End Root

B) x=2+5√3, 2−5√3
x equals 2 plus 5 Start Root 3 End Root comma 2 minus 5 Start Root 3 End Root

C) x=−2+3√5, −2−3√5
x equals negative 2 plus 3 Start Root 5 End Root comma negative 2 minus 3 Start Root 5 End Root

D) x=2+3√5, 2−3√5
x equals 2 plus 3 Start Root 5 End Root comma 2 minus 3 Start Root 5 End Root

Answer 1

To solve the quadratic equation \(x^2 + 4x = 41\) by completing the square, we can follow these steps:

1. Move the constant to the right side of the equation:
\[
x^2 + 4x - 41 = 0
\]
We rearrange it to:
\[
x^2 + 4x = 41
\]

2. Complete the square on the left side. To do this, take the coefficient of \(x\) (which is 4), divide it by 2 to get 2, and then square it to get 4. Add and subtract this square inside the equation:
\[
x^2 + 4x + 4 - 4 = 41
\]
This simplifies to:
\[
(x + 2)^2 - 4 = 41
\]

3. Move the constant term to the right:
\[
(x + 2)^2 = 41 + 4
\]
\[
(x + 2)^2 = 45
\]

4. Now, take the square root of both sides:
\[
x + 2 = \pm \sqrt{45}
\]

5. Simplifying \(\sqrt{45}\) gives:
\[
\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}
\]

6. So we have:
\[
x + 2 = \pm 3\sqrt{5}
\]

7. Finally, solve for \(x\):
\[
x = -2 + 3\sqrt{5} \quad \text{or} \quad x = -2 - 3\sqrt{5}
\]

The solutions to the equation \(x^2 + 4x = 41\) are:
\[
x = -2 + 3\sqrt{5}, \quad x = -2 - 3\sqrt{5}
\]

Therefore, the correct answer is:
**C) \(x = −2 + 3\sqrt{5}, −2 − 3\sqrt{5}\)**.