Solve by completing the square:x^2 + 6x + 1 = 0?

Question

x^2 + 6x + 1 = 0

x^2 + 6x + 1 -1 = 0 - 1 ( know I subtract 1 from both sides)

X^2 + 6x = -1

( now I'm lost)

Can you help me

Answer 1

This place does a great job explaining how to complete the square.  

http://www.purplemath.com/modules/sqrqua...

Answer 2

y=b^2-4ac

y=6^2-4*1*1=32

x1=(-b-sqy)/2a ----> x1=(-6-sq32)/2----->x1= -3-2sq2

x2=(-b+sqy)/2a ----> x1=(-6+sq32)/2----->x1= -3+2sq2

Answer 3

x² + 6x + 1 = 0

x² + 6x = -1

Take half of 6, square it then add it to both sides.

x² + 6x + 9 = -1 + 9

x² + 6x + 9 = 8

(x + 3)² = 8

x + 3 = ±√( 8 )

x = ±√( 8 ) - 3

x = ± 2√( 2 ) - 3

x ≈ -0.1715728753, x ≈ -5.828427125

Answer 4

since a quadratic is in form ax^2 +bx + c= y , then for it to be a square, c needs to equal (b/2)^2

for this problem, 6/2=3, and 3 squared is 9, so you would add 9 to both sides to make the quadratic: x^2 + 6x + 9 = 8

now factor: (x+3)^2=8

now solve for x: x+3= square root of 8

x= -3 plus/minus square root of 8  

Answer 5

x^2+6x+(6/2)^2=-1+(6/2)^2

x^2+6x+(3)^2=-1+(3)^2

x^2+6x+9=-1+9

(x+3)^2=8

x+3=(8)^(1/2)

x=(8)^(1/2)-3

Answer 6

x^2 +6x + --- = -1 + ____

You want to add half the b term squared to both sides:

x^2 +6x +9 = -1 +9

(x+3)^2 =8

x+3 =+/- sqrt 8 =+/-2sqrt 2

x=2sqrt2 -3 or -2sqrt2 -3