Complete the square of X^2 -2x+3?

Do as follows:

x² - 2x + 3
= x² - 2x + (2/2)² - (2/2)² + 3; dividing the coefficient in front of the x by 2, squaring it and then adding and substracting it.
= x² - 2x + 1 - 1 + 3; after simplifying
= (x² - 2x + 1) - 1 + 3; grouping the perfect square trinomial
= (x - 1)² + 2

Hope this helps!

(x² - 2x + 1) - 1 + 3
(x - 1)² + 2

x^2 - 2x + 3
= x^2 - 2x + 1+ 2
=(x-1)(x-1) + 2
=(x-1)^2 + 2

X^2 -2x+3
= x^2 - 2x + 1 + 2
= (x - 1)^2 + 2
the first term is a perfect sqaure

x^2 - 2x + 3
= (x - 1)^2 + 2

x^2-2x+1-1+3=(x-1)^2+2

x² - 2x + 3; by completing the squares, and assuming that the equation is set to zero (0), you get two (2) values of x which are;
x(+) = 1 + √2 i

and

x(-) = 1 - √2 i


Solution:

x² - 2x + 3 = (x² - 2x + 1) + 2

(x² - 2x + 1) is a complete square and if you get the factor, it becomes (x -1)(x -1) = (x - 1)²

Therefore, the given equation can now be rewritten as;

x² - 2x + 3 = (x - 1)² + 2

Assume that the equation can be equated to zero (0) to get the values of x;

(x - 1)² + 2 = 0

By transposition, you get

(x - 1)² = - 2
x -1 = ± √-2
x = 1 ± √-2

Note that √-2 = √2 * √-1, and √-1 = i, where i is an imaginary number

Hence,
x = 1 ± √2i

The positive (+) value of x is now known to be;
x(+) = 1 + √2 i

and the negative (-) value of x is

x(-) = 1 - √2 i