SOLUTION: please help?? it says solve by completing the square: x^2+8x+3=0 x^2+4x+1=0 x^2-5x+=11

Question 145099: please help??
it says solve by completing the square:
x^2+8x+3=0
x^2+4x+1=0
x^2-5x+=11
Found 2 solutions by stanbon, vleith:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
solve by completing the square:
x^2+8x+3=0
x^2+8x+16 = -3+16
(x+4)^2 = 13
x+4 = sqrt(13) or x+4 = -sqrt(13
x = -4+sqrt(13) or x = -4-sqrt(13
----------------
x^2+4x+1=0
x^2 + 4x + 4 = -1 + 4
(x+2)^2 = 3
x+2 = sqrt3 or x+2 = -sqrt3
x = -2+sqrt3 or x=-2-sqrt3
========================
and in the second answer is the negative sign in
or out of the sqare?
Ans: On both 1st and 2nd the negative refers to the square root of 13
and to the square root of 3.
------------------------------------
then on the last question x^2-5x=11 i didn't see anything i
don't know if it just didn't show up on my computor or you just didn't answer
that one??
Since I did #1 and #2 I left #3 for you to do.
-------------------
Cheers,
Stan H.

x^2-5x+=11

Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Read this -->http://www.purplemath.com/modules/sqrquad.htm
Now let's solve the first one using that method. The others are for you

Move the loose number over to the other side.

Divide through by whatever is multiplied on the squared term.
In this case that is 1, so we are done.
Take half of the coefficient (don't forget the sign!) of the x-term, and square it. Add this square to both sides of the equation.
=
=

Convert the left-hand side to squared form, and simplify the right-hand side. (This is where you use that sign that you kept track of earlier.)

Square-root both sides, remembering the "�" on the right-hand side. Simplify as necessary.

Solve for "x =".

and
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