How do you solve #x^2 + 6x + 2 = 0# using completing the square?

Question

Savannah Anderson · Accepted Answer

A square is allways of the form #x^2+2ax+a^2# Completing the square is you take half of the number #2a# with x and square it, which would be #x^2+6x+9#, but to even out with the #2# you've got, you write #2=9−7#: #x^2+6x+9−7=0→# #(x+3)^2−7=0→(x+3)^2=7→# #x+3=±√7→x_12=−3±√7#

Avery Alexander · Answer

undefined To solve the quadratic equation $x^2 + 6x + 2 = 0$ using completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $$x^2 + 6x = -2$$

2. To complete the square, take half of the coefficient of $x$, square it, and add it to both sides of the equation:
   $$x^2 + 6x + (6/2)^2 = -2 + (6/2)^2$$
   $$x^2 + 6x + 9 = -2 + 9$$

3. Simplify:
   $$x^2 + 6x + 9 = 7$$

4. Rewrite the left side as a perfect square:
   $$(x + 3)^2 = 7$$

5. Take the square root of both sides:
   $$x + 3 = \pm \sqrt{7}$$

6. Solve for $x$:
   $$x = -3 \pm \sqrt{7}$$

So, the solutions to the equation $x^2 + 6x + 2 = 0$ using completing the square are $x = -3 + \sqrt{7}$ and $x = -3 - \sqrt{7}$.