Question

Complete the square x^2+6x=7

Answer 1

\[x^2+6x-7=0\]
\[(x+7)(x-1)=0\]

Answer 2

x^2+6x =7
remember x^2 = x^2+2ax+a^2
x^2 is x^2. 6x is 2ax, we have to find a
we can find a by dividing 6x by 2, which is 3. So a =3
but remember a^2, so 3^2 is 9 (3 times 3 equals 9)
now you can plug that in.
x^2+6x+9=7
but what we do to one side we must do to the other
x^2+6x+9=7+9
right off the bat we can add 7 and 9
x^2+6x+9=16
ok now, lets factor it out by using difference of two perfect squares (you can ignore 16 for now)
(x+3)(x+3)
now you can bring back the 16
(x+3)^2=16 btw (x+3)^2 is same as (x+3)(x+3)
now, we have to find the square root of (x+3)^2 and 16
the square root of (x+3)^2 is just (x+3) and the square root of 16 is just 4
so
(x+3) = (+-) 4 remember the plus or minus sign with the square roots
now , subtract 3 from both sides
x= -3 (+-) 4
now just do basic arithmetic
x = -3 + 4
x= 1 is one solution
x = -3 - 4
x= -7 is another solution

Answer 3

http://www.wolframalpha.com/input/?i=x%5E2%2B6x%3D7

Answer 4

\[x^2+6x-7=0\]\[(x+3)^2-3^2-7=0\]\[(x+3)^2-16=0\]\[(x+3)^2=16\]\[x+3=\pm 4\]\[x=1 , -7\]