How to solve the quadratic equation by completing the square

I'm taking exams and I really want to pass this so I can be a senior...so how can I solve please make it neat.. Thx.


x^2-2x+4=0

My teacher gave me the answer and it's::
1+- i{square root:3}
how did she get that answer??
I'm confused because the imaginary number is 3


PLEASE HELP

Welp,

x^2 -2x +4 = 0

x^2-2x=-4

x^2-2x+(1)=-4+(1)

(x-1)^2=-3

√((x-1)^2)=√(-3)

x-1=i√3 x-1=-i√3

x=1+1√3, x=1-i√3

Yep.

To complete the square you have to intuitively find a square function within your function and kind of separate it from the rest.

For this one you look at all the terms that have x in them and find the square function that would give those terms with x.

x^2-2x+4=0 you can separate the square function (x-1)^2= x^2-2x+1

*note that in the square function the x terms are the same as the ones in your given function, thats good because that's what you want.

So now that square function only differs from your given function by 3
Therefore you can rewrite your function as x^2-2x+1+3=0= (x-1)^2+3

Now isolate the x term

(x-1)^2=-3

Take the square root and simplify

x= 1 +/- sqrt(-3)

Factor out the i= sqrt(-1)

and...

x= 1 +/- i*sqrt(3)

Subtract 4 from both sides

x^2 - 2x = -4

Add 1 to both sides (b/2)^2

x^2 - 2x + 1 = -3

(x - 1)^2 = -3

x - 1 = +/- i(sqrt3)

x = -1 +/- i(sqrt3)

I hope this information was very helpful.

x^2 - 2x + 4 = 0
x^2 - 2x = -4
x^2 - 2x + 1 = -4 + 1
(x - 1)^2 = -3
x - 1 = +/- sqrt(-3)
x = 1 +/- i * sqrt(3)