Complete the square to solve the following quadratic equation: x^2+4x=41

Question

.(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) \(x = −2 + 3\sqrt{5}, −2 − 3\sqrt{5}\).