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

x^2 + 6x + 2 = 0
x^2 + 6x = -2
x^2 + 6x + (3^2) = -2 + 3^2
x^2 + 6x + 9 = 7
(x + 3)^2 = 7
x = +/- sqrt(7) - 3

On the third step, remember to add to the whole equation the *square of the half the coefficient of the x term*.

X 2 6x 2

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

Question Number 1 :
For this equation x^2 + 6*x + 2 = 0 , answer the following questions :
A. Use completing the square to find the root of the equation !

Answer Number 1 :
The equation x^2 + 6*x + 2 = 0 is already in a*x^2+b*x+c=0 form.
By matching the constant position, we can derive that the value of a = 1, b = 6, c = 2.

1A. Use completing the square to find the root of the equation !
x^2 + 6*x + 2 = 0 ,divide both side with 1
So we get x^2 + 6*x + 2 = 0 ,
And the coefficient of x is 6
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = 6/2 = 3
Which means we can turn the equation into x^2 + 6*x + 9 - 7 = 0
Which can be turned into ( x + 3 )^2 - 7 = 0
Which is the same with (( x + 3 ) - 2.64575131106459 ) * (( x + 3 ) + 2.64575131106459 ) = 0
Which is the same with ( x + 3 - 2.64575131106459 ) * ( x + 3 + 2.64575131106459 ) = 0
Just add up the constants in each brackets, and we get ( x + 0.354248688935409 ) * ( x + 5.64575131106459 ) = 0
We get following answers x1 = -0.354248688935409 and x2 = -5.64575131106459

x^2+6x+2 = 0
ADDING AND SUBTRACTING THE SQUARE OF THE HALF OF THE COEFFICIENT OF x
SO WE HAVE ,,,,,,,,
1/2 (6) = 3
3^2 = 9
NOW ,,,,,
x^2+6x+2+9 - 9 = 0
x^2+6x+9 - 7 = 0
x^2+6x+9 = 7
x^2+3x+3x+9 = 7
x ( x+3 ) + 3 ( x+3 ) = 7
(x+3)(x+3) = 7
(x+3)^2 = 7
x+3 = ROOT OF 7
THEREFORE ,,,,,
X = -3 + ROOT OF 7

x^2+6x+2=0
(-2 both sides)

x^2+6x=-2
(Take half of b [ax^2+bx+c=0], in this case 6, and square it)

6/2=3 3^2=9
(add the result, in this case 9, to both sides)

x^2+6x+9=7
(the left side is now a perfect square and can be abbreviated)

(x+3)^2 = 7
(from here, just solve for the square)

x+3= +/- rt7
(subtract 3 from both sides)

x = -3 +/- rt7
(rt7 = the square root of seven. I don't know where to get the square root symbol on a keyboard =P)

And you're done =)

Here are the steps to apply to any completing the square problem:

Given ax² + bx + c = 0

1. Move the constant, c, to the right side:

x² + 6x = -2

2. Make sure the coefficient, a, of x² is 1. Here, it is.
.....If not, divide every term by a.

3. Take 1/2 the value of b and square it. Add the result to both sides.

b = +6 (make sure to include the sign; if negative, keep negative)
1/2 b = +3 (you will see the +3 in the factored form as well)
b² = 9 .·. add 9 to both sides

x² + 6x + 9 = -2 + 9
x² + 6x + 9 = 7

4. Rewrite as a perfect square:
(x + 3)² = 7 ......... Note the +3 in the factored form.
............................ It will always factor to (x + b/2)²

5. Take the square root of both sides - don't forget ±!
x + 3 = ±√7

6. Solve for x by subtracting 3 from both sides.
x = -3 ± √7

Done!

x^2+6x+2 +9 -9 = 0

x^2+6x+9=7
(x+3)^2=7

x^2 + 6x + 2 = 0
x^2+6x=-2
x^2+6x+9=-2+9
(x+3)^2=7
x+3=±√7
x=-3±√7

x=-3+√7, -3-√7 answer//

x^2+6x+2=0

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

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

x^2+6x+9=7

(x+3)^2= 7

take square root of both sides

x+3= + or - square root of 7

x=-3+or-square root of 7