Question

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

Answer 1

`x^2+3x+1=(x+3/2)^2-9/4+1=(x+3/2)^2-5/4`

Now remember the identity `(a+b)(a-b)=a^2-b^2`

In our problem we write (because is the same) `(x+3/2)^2-sqrt5^2/2^2` applying identity with `a=x+3/2` and `b=sqrt5/2` and we have

`(x+3/2+sqrt5/2)(x+3/2-sqrt5/2)` or the equivalent

`(x+(3+sqrt5)/2)(x+(3-sqrt5)/2)`