How do I find all real and complex zeros of #x^2-4x+7#?

1 Answer
Oct 31, 2015

Answer:

Use the quadratic formula to find zeros occur for:

#x = 2+-i sqrt(3)#

Explanation:

The quickest way is probably to use the quadratic formula.

#x^2-4x+7# is in the form #ax^2+bx+c#, with #a=1#, #b=-4# and #c=7#.

So its zeros occur for:

#x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-4)+-sqrt((-4)^2-(4xx1xx7)))/(2xx1)#

#=(4+-sqrt(16-28))/2 = (4+-sqrt(-12))/2 = (4+-i sqrt(12))/2#

#=(4+-i*2sqrt(3))/2 = 2+-i sqrt(3)#

Alternatively, you can complete the square as follows:

#0 = x^2-4x+7 = x^2-4x+4+3 = (x-2)^2+3#

Subtract #3# from both ends to get:

#(x-2)^2 = -3#

So:

#x-2 = +-sqrt(-3) = +-i sqrt(3)#

Add #2# to both sides to get:

#x = 2 +-i sqrt(3)#