How do I complete the square if the coefficient of #x^2# is not 1?

Question

James Allen · Accepted Answer

Recall that #(ax + b)^2 = a^2x^2 + 2abx + b^2# So the general idea to get #b^2# is by getting #x#'s coefficient,
dividing by #2a#, and squaring the result. Example #3x^2 + 12x# #a = 3#
#2ab = 12#
#6b = 12#
#b = 2#
#3x^2 + 12x + 4 = (3x + 2)^2# You can also factor out #x^2#'s coefficient.. and proceed with completing the square. Example:
#2X^2 +4X# #2(X^2 + 2X)# #2(X + 1)^2#

Greyson Brown · Answer

undefined To complete the square when the coefficient of $ x^2 $ is not 1, follow these steps:

1. Divide the entire equation by the coefficient of $ x^2 $ to make the coefficient of $ x^2 $ equal to 1.
2. Rewrite the equation in the form $ x^2 + bx + c $.
3. Take half of the coefficient of $ x $ (b/2) and square it to get $ (b/2)^2 $.
4. Add $ (b/2)^2 $ to both sides of the equation.
5. Rewrite the equation as a perfect square trinomial on the left side.
6. Factor the perfect square trinomial and simplify the right side of the equation.
7. Solve for x by taking the square root of both sides and isolating x.

This process allows you to rewrite the quadratic equation in vertex form, making it easier to solve by completing the square.