How do you solve the quadratic #2x^2-6x+5=0# with complex numbers?
1 Answer
Dec 15, 2017
Explanation:
#0 = 2(2x^2-6x+5)#
#color(white)(0) = 4x^2-12x+10#
#color(white)(0) = (2x)^2-2(2x)(3)+3^2+1#
#color(white)(0) = (2x-3)^2+1^2#
#color(white)(0) = (2x-3)^2-i^2#
#color(white)(0) = ((2x-3)-i)((2x-3)+i)#
#color(white)(0) = (2x-3-i)(2x-3+i)#
So:
#2x = 3+-i#
and:
#x = 3/2+-1/2i#
