Question

Complete the Square: x^2 + 5x = -6
I don't understand this assignment of completing the squares

Answer 1

\[x^2+5x+6=0 \]\[x^2 + 2x+3x+6=0\]\[x(x+2)+3(x+2)=0\]\[x=-2 , -3\]

Answer 2

okay...thanks @kamalhandoo @Hero

Answer 3

see http://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/completing-the-square

Answer 4

@Hero because the problem tells him to complete the square!

\[x^2 + 5x = -6\]To complete the square, we need the coefficient of the x^2 to be 1, and we've got that. Otherwise, we would divide the whole equation by the coefficient. We move the x^2 and x terms to the left, and the numeric remainder to the right. We've got that, too.

Next, we take half of the coefficient of the x term, square it, and add it to both sides.
\[x^2 + 5x + (\frac{5}{2})^2 = -6 + (\frac{5}{2})^2\]Now we can write the left hand side as a square:\[(x+\frac{5}{2})^2 = (\frac{5}{2})^2 - 6\]or\[(x+\frac{5}{2})^2 = \frac{25}{4}-\frac{24}{4} = \frac{1}{4}\]
Take the square root of each side and solve for \(x\):

Root #1:
\[x+\frac{5}{2} = \frac{1}{2}\]
Solve for \(x\)
Root #2:
\[x + \frac{5}{2} = -\frac{1}{2}\]
Solve for \(x\)
(remember that \(\sqrt{(1/4)} = -1/2\) as well as \(1/2\)).