How do you solve #x^2 + 4x - 12 = 0# by completing the square?

Question

Abigail Allen · Accepted Answer

The solutions are #color(green)(x = 2# ,# color(green)(x = -6#

#x^2 + 4x - 12 = 0 #

#x^2 + 4x = 12#

We add 4 to both sides in order to write the Left Hand Side as a Perfect Square:

#x^2 + 4x + 4 = 12 + 4#

#x^2 + 2 * x * 2 + 2^2 = 16#

Using the Identity #color(blue)((a+b)^2 = a^2 + 2ab + b^2#, we get #(x+2)^2 = 16#

#x + 2 = sqrt16# or #x +2 = -sqrt16#

#x + 2 = 4# or #x +2 = -4#

#x = 4 -2 # or #x = -4 -2 #

#color(green)(x = 2# ,# color(green)(x = -6#

Eva Abramson · Answer

undefined To solve the quadratic equation $ x^2 + 4x - 12 = 0 $ by completing the square, follow these steps:

1. Move the constant term to the other side of the equation:
   $ x^2 + 4x = 12 $

2. Take half of the coefficient of $ x $ (which is 4), square it, and add it to both sides of the equation:
   $ x^2 + 4x + (4/2)^2 = 12 + (4/2)^2 $
   $ x^2 + 4x + 4 = 12 + 4 $

3. Simplify both sides:
   $ x^2 + 4x + 4 = 16 $

4. Rewrite the left side as a perfect square:
   $ (x + 2)^2 = 16 $

5. Take the square root of both sides:
   $ x + 2 = \pm \sqrt{16} $

6. Solve for $ x $:
   $ x + 2 = \pm 4 $
   $ x = -2 \pm 4 $

7. Solve for the two possible values of $ x $:
   $ x_1 = -2 + 4 = 2 $
   $ x_2 = -2 - 4 = -6 $

Therefore, the solutions to the equation $ x^2 + 4x - 12 = 0 $ are $ x = 2 $ and $ x = -6 $.