Quadratics.

Question

Quadratics.

compassionate · Answer

Excuse me, what happened in step 3?

luigi0210 · Answer

They took half the number, $b$ term in $Ax^2+Bx+C$, and squared it.

luigi0210 · Answer

Also, what you do to one side you have to do to the other~

compassionate · Answer

I have no idea what b is.

luigi0210 · Answer

In a quadratic, it's usually just the term that has a variable that isn't square. So $x^2-4x$, the $b$ is $-4$.

johnweldon1993 · Answer

So they had 2 quadratics, the first being

$$\large (x^2 + 4x)$$

They completed the square here...you know a quadratic looks like $\large Ax^2 + Bx + C$

Well....we have the $\large A = 1$ since the coefficient of the x^2 is 1
we have that $\large B = 4$ since the coefficient of the 'x' is 4

So they completed the square...by taking half that B...and then squaring the result...and adding that to both sides

johnweldon1993 · Answer

Sorry, that's supposed to be $\large -4x$ but regardless the process is the same...just replace 4 with -4 in my explanation :)

luigi0210 · Answer

Either way, negative or positive, the square always gives you a positive~

compassionate · Answer

Half of b would be 2... But I don't see that... I see on the left we have 4x+4, and where does that 1 in the -2y+1?

compassionate · Answer

I have absolutely no idea what's going on at this point. 

Explain to me like I'm completely new to quadratics. The only thing I know about quadratics is the form...

compassionate · Answer

Okay, but if we took half of 4, why are we adding for to the first one?

luigi0210 · Answer

Comp, OS is too whacky right now ._.

compassionate · Answer

I know. IT's freaking out on me...