How do you solve #2x^2-3x+1=0# by completing the square?

1 Answer
George C.
Dec 22, 2016

#x=1# or #x=1/2#

Explanation:

The difference of squares identity can be written:

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

We will use this below with #a=(4x-3)# and #b=1#.

I prefer not to have to do much arithmetic involving fractions, so I would pre-multiply this equation by #8# to avoid them and get:

#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#

#color(white)()#
Footnote

Why #8#?

#8 = 2*2^2#

The first factor of #2# makes the leading term into a perfect square. The additional #2^2# factor avoids us having to divide #3# by #2# and end up working with #1/2#'s and #1/4#'s

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