Question

How do you factor and solve `x^2+3x+1=0`?

Answer 1

`x = -3/2+sqrt(5)/2` or `x = -3/2-sqrt(5)/2`

Explanation:

Use the difference of squares identity:

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

with `a=(x+3/2)` and `b=sqrt(5)/2` as follows:

`0 = x^2+3x+1`

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

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

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

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

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

Hence:

`x = -3/2+sqrt(5)/2` or `x = -3/2-sqrt(5)/2`