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:

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

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