consider the equation x^2+12x+c=0

Question

if c= -28, list the steps to solve the equation above by completing the square, and give the solution or solutions. 

Answer 1

First, remember that (x+a)² = x² +2ax + a² so the first step in completing the square is to realize that the binomial you want is (x + (b/2)) which, when squared gives x² + bx + b²/4.

 

Therefore you need to subtract the b²/4 from the expression later on.

In the current example you will want (x + 6)² 

and you will then need to solve:

 

(x + 6)² - 36 -28 = 0 => (x + 6)²  = 64

Remember when you completed the square you added 36 which now must be subtracted.

 

then (x + 6) = ±8

 

The solutions are x = 2 and x = -14 

 

Just for practice you might want to develop the quadratic formula which comes about by completing the square of the generalized quadratic: Ax² + Bx + C = 0.

Answer 2

x^2+12x-28=0

1. 12/2=6

2. (x+6)^2=x^2+12x+36

3. If we subtract 64 from this expression it becomes

x^2+12x+36-64=x^2+12x-28 and so equals 0

We now have (x+6)^2-64=0

This is our old friend a^2-b^2=(a+b)(a-b)

(x+6)^2-8^2=(x+6+8)(x+6-8)=(x+14)(x-2)=0 and so x=-14 or x=2