Question

How do you complete a square?

Answer 1

Ha ha.. in algebra

Answer 2

omg that was so funny

Answer 3

that was a good one though :)

Answer 4

Do you have an example equation to work from?

Answer 5

if the side in a then the squares side is 4a

Answer 6

I guess we can do \(ax^2 + bx + c\) but that sometimes gets ugly.

Answer 7

x^2+6x+7=0

Answer 8

The idea of completing the square, as you might know, is to get rid of that middle x, so the whole thing can be written as \((a + h)^2 + k = 0\)

Answer 9

okay so what is the first step?

Answer 10

Ignore the 7 for now. Think about what would make \(x^2 + 6x + ???\) something that can be written like \((x + a)^2\). The answer is 9, course. So we add 9 to both sides of the equation.
\[x^2 + 6x + 9 + 7 = 0 + 9\]
Then we can clean that right up.
\[(x+3)^2 + 7 = 9 \\ (x+3)^2 - 2 = 0\]

Answer 11

ok, that makes sense, then what?

Answer 12

If you want to solve it, then just move the free number over to the other side and take the square root.
\[(x + 3)^2 - 2 = 0 \\ (x+3)^2 = 2 \\ \sqrt{(x+3)^2} = \pm\sqrt{2}\]
So then we have
\(x + 3 = \sqrt{2}\) and \(x + 3 = -\sqrt{2}\)
So our two solutions are \(-3 + \sqrt{2}\) and \(-3 - \sqrt{2}\)

Answer 13

wow, thank you! I totally get it now

Answer 14

If you want to do it for the generic \(ax^2 + bx + c\) we can do that, too.
Amusingly (and yet predictably) you'll end up with the quadratic formula.

Answer 15

I always remember take half of the coefficient of the x term and square it.