How do you complete the square?

Question

Argh, I have a math test tomorrow on completeing the square, but I totally forgot how to do it. Can someone just show me an example? 10 points for best answer~ Please walk me through the steps.

Question:
x^2 - 8x +22 = 0

The question is right. My teacher gave us some sheet to practice. Thank you everyone for the help!

Answer 1

x^2 - 8x + 22 = 0

The main step is to add and subtract "half-squared" of the coefficient of x. In this case, the coefficient of x is -8; half of -8 is -4, squared is 16. Add and subtract 16 in this manner.

x^2 - 8x + ? + 22 - ? = 0

x^2 - 8x + 16 + 22 - 16 = 0

Note that the equation does not change, because the +16 and -16 offset each other. With that said, we now factor the now-square trinomial.

(x - 4)^2 + 22 - 16 = 0

(x - 4)^2 + 6 = 0

(x - 4)^2 = -6

Taking the square root of both sides,

x - 4 = +/- sqrt(6) i

x = 4 +/- sqrt(6)i

Answer 2

x^2 - 8x +22 = 0
transpose constant term to the other side
x^2 - 8x = -22
add on both side square of the half of the coefficients
x^2 - 8x +16 = -22+16 = -6
LHS is a perfect square but RHS is negative, therefore it can be written as
(x-4)^2 = (6^1/2 i^2)^2 where i^2 = -1

x = 4+isqrt(6), x = 4-isqrt(6)

x = 4+2.449i , x = 4 - 2.449i

Answer 3

( x ² - 8 x + 16 ) + 22 = 16
( x - 4 ) ² = - 6
( x - 4 ) ² = 6 i ²
x - 4 = ± i √ 6
x = 4 ± i √ 6

Answer 4

i agree with Puggy 's answer
but the thing is ..
im just wondering if the question is right..
Puggy's last few steps shows that (x-4)^2 = -6
it cant be solved..
everything under square root have to be > 0

Answer 5

x²-8x+22=0
x²-8x+(4)²-(4)²+22=0
(x+4)²-16+22=0
(x+4)²+6=0

there you have it.

Answer 6

ax^2 + bx+c=0
if S=b^2-4ac <0 => "vn"
if S= 0 so x= - b/2a
if S>0 =>
x1=(-b+sqrt(s))/2a
x2=(-b-sqrt(s))/2a