What are the roots of this equation?
x²-4x+9=0

a. 1±i√5

b. 2±i√5

c. 2±i√3

d. 1±i√52

x = 2 ± (i)√5  (Answer b)

Step-by-step explanation:

x²-4x+9=0 can be solved in a variety of ways; the first two that come to mind that are also appropriate are (1) completing the square and (2) using the quadratic formula.

Completing the square is fast here:

Rewrite x²-4x+9=0 as   x²-4x                    +9=0

Identify the coefficient of the x term:  it is -4

Take half of that, obtaining -2

Square this result, obtaining 4

Add 4 to   x²-4x                    +9=0, in the blank space in the middle, and then subtract 4:      x²-4x +4          -4         +9=0

Rewrite           x²-4x +4 as the square of a binomial:

                          (x - 2)² - 4 + 9  = 0  →    (x - 2)²  = -5

Take the square root of both sides:       x - 2  = ±√(-5)  =  ± (i)√5

Then x = 2 ± (i)√5

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .