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

1 Answer
Jan 8, 2017

Answer:

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

Explanation:

#f(x) = 2x^2-3x+1#

The difference of squares identity can be written:

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

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

First pre-multiply by #8# to avoid the need to do arithmetic with fractions:

#0 = 8f(x)#

#color(white)(0) = 8(2x^2-3x+1)#

#color(white)(0) = 16x^2-24x+8#

#color(white)(0) = (4x)^2-2(4x)(3)+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)#

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