Question

How do I solve `4x^2 - 6x + 1 = 0` by completing the square?

Answer 1

`4x^2-6x=-1`

Factor out a `4`

`4(x^2-6/4x)=-1`

Simplify fraction

`4(x^2-3/2x)=-1`

Take half of x term and square the result

`((-3/2)/2)^2=((-3/2)*(1/2))^2=(-3/4)^2=9/16`

Add `4*9/16` or `9/4` to the right side because we factored out a `4`.
Add `9/16` to the left side.

`4(x^2-3/2x+9/16)=-1+9/4`

`4(x-3/4)^2=-1+9/4`

`4(x-3/4)^2=-4/4+9/4`, common denominators `-1=-4/4`

`4(x-3/4)^2=5/4`

`(x-3/4)^2=(5/4)/4`

`(x-3/4)^2=5/16`

`sqrt((x-3/4)^2)=+-sqrt(5/16)`

`x-3/4=+-sqrt(5/16)`

`x-3/4=+-sqrt(5)/4`

`x=+-sqrt(5)/4+3/4`

`x=(+-sqrt(5)+3)/4`