What are the roots of the equation x^2-4x+3=0. Work on two methods.

The first method is to use
factorization:

The original equation is:
x^2-4x+3=0

We'll write 3 as the difference 4 - 1 =
3

We'll re-write the equation, changing 3 with the
difference:

x^2-4x+4-1=0

Now,
we'll combine the first and the last terms and the middle
terms:

(x^2 - 1) - (4x - 4) =
0

We'll write the difference of squares as a
product:

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

We'll factorize (4x - 4) by
4:

(4x - 4) = 4(x-1)

Now,
we'll re-write the equation:

(x-1)(x+1) - 4(x-1) =
0

We'll factorize by (x-1) and we'll
get:

(x-1)(x+1-4) = 0

We'll
set each factor as zero:

x - 1 =
0

We'll add 1 both sideS:

x =
1

x - 3 = 0

We'll add 3 both
sides:

x = 3

The
roots of the equation are: {1 ;
3}. 

The second method is to
apply the quadratic formula:

x1 =
[-b+sqrt(b^2-4ac)]/2a

x1 =
[4+sqrt(16-12)]/2

x1 =
(4+2)/2

x1 =
3

x2 =
(4-2)/2

x2 =
1

As we can see, we've get the
same result!