How do you solve #x^2+3x+1=0#?

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Answer 1 · 2015-05-29T19:49:17

We can solve this by completing the square ...

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

#= x^2+3x+9/4 - 9/4 + 1#

#=(x+3/2)^2-5/4#

Add #5/4# to both ends to get:

#(x+3/2)^2 = 5/4#

So

#x+3/2 = +-sqrt(5/4) = +-sqrt(5)/sqrt(4) = +-sqrt(5)/2#

Subtract #3/2# from both ends to get:

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