Question

solve for the complete square
x^2+4x-8=0

Answer 1

\[\large x^2+4x-8=0\]


\[\large x^2+4x=8\]


\[\large x^2+4x+4=8+4\]


\[\large (x+2)^2=12\]


\[\large x+2=\pm\sqrt{12}\]


\[\large x+2=\pm2\sqrt{3}\]


\[\large x=-2\pm2\sqrt{3}\]

Answer 2

b/2^2 wasnt used..

Answer 3

that's the point where I added 4 to both sides

Answer 4

take half of 4 (the x coefficient) and square it to get 4, then add it to both sides

Answer 5

okay i understand thanks

Answer 6

We are trying to get left hand side to be in the form \[(x + a)^2 = c\] or \[x^2 + 2ax + a^2 = c\]

\[2a=4, a=2\]
\[(x+2)^2 = x^2 + 4x + 4\]

So we need to add 12 from both sides to make the left side equal to (x+2)^2, giving us \[(x+2)^2 = 12\] which means \[(x+2) = \pm\sqrt(12)\] and
\[x=\pm2\sqrt(3)-2\]