What are the solutions of x^2-2x+26=0?

x = 1 + (i)5 and x = 1 - (i)5

Step-by-step explanation:

x^2-2x+26=0 can be rewritten by completing the square of x^2-2x, as follows:

x^2-2x+26=0

x^2-2x+ 1  - 1  +26=0 (this 1 comes from halving the coefficient of x (which is -2), obtaining -1, squaring the result, and then adding this 1 to and subtracting this 1 from x^2-2x)  →  x^2-2x+ 1  - 1  +26=0

Rewriting x^2-2x+ 1 as the square of a binomial, we get:

 (x - 1)²  - 1  +26=0, or  (x - 1)²  - 1  +26=0, or   (x - 1)² = -25

Taking the square root of both sides yields x - 1 = ±(i)5.

Thus, the roots are x = 1 + (i)5 and x = 1 - (i)5  (Answer C)

a. is correct

hi! i think the answer is 1/2 : )