Question

Complete the square to solve the following quadratic equation: x6^2+2x-8=0

A. x=-10,x=8
B. x=-2,x=4
C. x=10,x=-8
D. x=2,x=-4

Answer 1

To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, we can follow these steps:

1. Move the constant term to the other side of the equation:
\[
x^2 + 2x = 8
\]

2. Take half of the coefficient of \( x \), square it, and add it to both sides. The coefficient of \( x \) is 2, so half of it is 1, and squaring it gives us \( 1^2 = 1 \).
\[
x^2 + 2x + 1 = 8 + 1
\]
\[
x^2 + 2x + 1 = 9
\]

3. Rewrite the left side as a squared term:
\[
(x + 1)^2 = 9
\]

4. Take the square root of both sides:
\[
x + 1 = \pm 3
\]

5. Solve for \( x \):
- If \( x + 1 = 3 \):
\[
x = 3 - 1 = 2
\]
- If \( x + 1 = -3 \):
\[
x = -3 - 1 = -4
\]

Thus, the solutions are \( x = 2 \) and \( x = -4 \).

The correct answer is:
**D. \( x = 2, x = -4 \)**