Question

How do you solve using the completing the square method `2x^2-8x+3=0`?

Answer 1

`x = 2+-sqrt(10)/2`

Explanation:

Complete the square then use the difference of squares identity:

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

with `a=(2x-4)` and `b=sqrt(10)` as follows.

Multiply by `2` first to make the leading term a perfect square:

`0 = 2(2x^2-8x+3)`

`=4x^2-16x+6`

`=(2x)^2-2(2x)(4)+6`

`=(2x-4)^2-16+6`

`=(2x-4)^2-10`

`=(2x-4)^2-(sqrt(10))^2`

`=((2x-4)-sqrt(10))((2x-4)+sqrt(10))`

`=(2x-4-sqrt(10))(2x-4+sqrt(10))`

`=(2(x-2-sqrt(10)/2))(2(x-2+sqrt(10)/2))`

`=4(x-2-sqrt(10)/2)(x-2+sqrt(10)/2)`

Hence:

`x = 2+-sqrt(10)/2`