Question

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

Answer 1

`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