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

Question

2084 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-08T10:01:52

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

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

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

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