Completing the square?

Question

x^2+ 6x +1 =0

solve by completing the square! I tried over and over and cant figure out how to work this using that specific method please explain!

Answer 1

Relax. If you have (x+a)^2= x^2 +2ax+a^2, you will note that by comparison of like-power terms,
(1) The last term of the expanded square will be a^2 AND
(2) The middle term of the expanded square will be 2a.
So from (2): 2a= 6 and a=3.
So you can form the square TAKING CARE TO BALANCE the difference between what goes into the square and what you started with for the constant term. THUS
x^2+6x+1 = x^2+6x+9 - 8 = (x+3)^2-8.....
And (x+3)^2= 8. Taking sqrts,
x+3 = +/- sqrt(8) or +/- 2*sqrt(2).
So x = -3 +/- 2*sqrt(2)

Answer 2

hold on ... this will take a minute.
Okay, you need to figure out which perfect square it looks like. That would be (x+3)^2 = x^2 + 6x + 9.
Now take the 1 to the other side of the equation because you don't need it there.
x^2 + 6x = -1
Now add 9 to both sides to make it a perfect square.
x^2 + 6x + 9 = -1 + 9 = 8
you can rewrite
(x+3)^2 = 8
Now square root both sides.
x+3 = +- sqrt8 or +- 2root2
x = -3 +- 2root2

Answer 3

x^2 + 6x + 1 = 0

x^2 + 2(1)(3)x + 3^2 - 3^2 + 1 = 0

(x + 3)^2 -8 =0

(x + 3)^2 = 4(2)

(x + 3)^2 = [2sqrt(2)]^2

x + 3 = 2 sqrt(2) or - 2sqrt(2)

x = 2sqrt(2) - 3 or

= - 2sqrt(2) -3

Answer 4

you need to use the quadratic equation to solve it