How do you solve #2x^2-3x+1=0# by completing the square?
1 Answer
Explanation:
The difference of squares identity can be written:
#a^2-b^2 = (a-b)(a+b)#
We will use this below with
I prefer not to have to do much arithmetic involving fractions, so I would pre-multiply this equation by
#0 = 8(2x^2-3x+1)#
#color(white)(0) = 16x^2-24x+8#
#color(white)(0) = 16x^2-24x+9-1#
#color(white)(0) = (4x-3)^2-1^2#
#color(white)(0) = ((4x-3)-1)((4x-3)+1)#
#color(white)(0) = (4x-4)(4x-2)#
#color(white)(0) = (4(x-1))(2(2x-1))#
#color(white)(0) = 8(x-1)(2x-1)#
Hence:
#x = 1" "# or#" "x = 1/2#
Footnote
Why
#8 = 2*2^2#
The first factor of