How do you solve #2x^2 - 6x +1 = 0# by completing the square?
1 Answer
Jan 21, 2017
#x=sqrt(7/4)+3/2#
#x=-sqrt(7/4)+3/2#
Explanation:
Given -
#2x^2-6x+1=0#
#2x^2-6x=-1#
#(2x^2)/2-(6x)/2=(-1)/2#
#x^2-3x=(-1)/2#
#x^2-3x+9/4=(-1)/2+9/4#
#(x-3/2)^2=(-2+9)/4=7/4#
#(x-3/2)=+-sqrt(7/4)#
#x=sqrt(7/4)+3/2#
#x=-sqrt(7/4)+3/2#
