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

##### 1 Answer

#### 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