How do you solve the quadratic #u^2-2u+3=0# using any method?

1 Answer
Sep 26, 2016

Answer:

#u = 1+-sqrt(2)i#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

Use this with #a=(u-1)# and #b=sqrt(2)i# as follows:

#0 = u^2-2u+3#

#color(white)(0) = u^2-2u+1+2#

#color(white)(0) = (u-1)^2-(sqrt(2)i)^2#

#color(white)(0) = ((u-1)-sqrt(2)i)((u-1)+sqrt(2)i)#

#color(white)(0) = (u-1-sqrt(2)i)(u-1+sqrt(2)i)#

Hence #u = 1+-sqrt(2)i#