Is this a correct derivation of completing the square?

Question

$x^2 + bx$

$=x^2 + bx + c - c$

$=(x + k)^2 - c$

$=x^2 + 2kx + (k^2 - c) = x^2 + bx + 0$

This implies:

$2k = b$, so $k = b/2$, and:

$k^2 - c = 0$, or $k^2 = c$, or $(b/2)^2 = c$

So to complete the square we are making the transformation:

$x^2 + bx \implies (x + b/2)^2 - (b/2)^2$

Answer 1

Your approach is just fine. This is often how mathematics is done; in the initial stages, you may run into a dead end or use unnecessary variables. Later on, once you have found that $k$ does exist and equals $b/2$, you can rework your exposition to avoid $k$ entirely and just use $b/2$. You will often see proofs in textbooks where the author "magically" knows what substitutions to use, etc.; this comes from experience and from working through the problem as you have done, then presenting it as if it were obvious all along.

That said, there's a circularity in your first few lines that is confusing and should be avoided. You wrote:

$$\begin{align} x^2+bx &= x^2+bx+c−c\\ &=(x+k)^2−c\\ &=x^2+2kx+(k^2−c)\\ &=x^2+bx+0 \end{align}$$

It's strange to begin and end with $x^2+bx$, since it isn't clear where you're going. Instead, leave off that last line, thus:

$$\begin{align} x^2+bx &= x^2+bx+c−c\\ &=(x+k)^2−c\\ &=x^2+2kx+(k^2−c) \end{align}$$ for some $k$. Equating coefficients, we have $2k=b$ and $k^2-c=0$. [etc.]

Answer 2

It's true, but you can get it much more easier: $$x^2+bx=x^2+bx+\frac{b^2}{4}-\frac{b^2}{4}=\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$$

Answer 3

Yes it is correct, indeed

$$(x + b/2)^2 - (b/2)^2=x^2+bx+\frac{b^2}4-\frac{b^2}4-=x^2+bx$$

Answer 4

Okay. You are trying to derive.

As you want to derive $x^2 + bx = (x + k)^2 + c$ I'd start with that line:

"we want $x^2 + bx = (x+k)^2 +c$ and we need to find the $c$ and $k$ that will make it so.

You started instead with

$x^2 + bx = x^2 + bx + c - c = (x+k)^2 -c$ with, I guess (????) then assumption that $(x+k)^2 = x^2 + bx + c$. But without any reason I don't get what you are doing. What is that $c$. It just came out of nowhere. Why do you think that $x + bx + c = (x + k)^2$ where did the $k$ come from? What are you doing?

It doesnt seem clear.

As you want to solve $x^2 + bx = (x + k)^2 -c$ then start there. It's almost as though you are afraid to ask the right question so you ask the wrong question $x^2 + bx + c -c = (x + k)^2 - c$.

But if you say:

"$x^2 + bx = (x+k)^2 -c $; what are the $k$ and $c$ we need to make the happen?"

The solution is clear.

$(x+k)^2 - c = x^2 + 2kx + k^2 -c$ and we want $x^2 + bx = x^2 + 2kx +k^2 - c$ so $k = \frac b2$ and $k^2 - c = \frac {b^2}4 - c = 0$. So $c = -\frac {b^2}4$.