How do I solve #(x-2)^2 = -3# using the square root property?

1 Answer
Aug 21, 2015

Answer:

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

Explanation:

If #a^2 = b#
then #a = +-sqrt(b)#

In this case #a = (x-2)# and #b = -3#
So
#color(white)("XXXX")x-2 = +-sqrt(-3)#

Note that #sqrt(-3) = sqrt(3)*sqrt(-1)#
and there is no Real value #= sqrt(-1)#.
However, if we are allowed to use Complex values
by definition #i =sqrt(-1)#

Therefore, we can write
#color(white)("XXXX")x-2 =+- (sqrt(3))i#
and after adding 2 to both sides
#color(white)("XXXX")x = 2+-(sqrt(3))i#