# How do I get from $-4(x+\frac{1}{2})^2+4$ to $4-(2x+1)^2$?

I'm trying to follow along with printed solutions to previous exam questions, and I'm completely stumped as to how this makes any sense.

I had to complete the square for the function $3-4x-4x^2$, and I got $-4(x+\frac{1}{2})^2+4$. But that's not the final form I need it in because I"m trying to use the solution to integrate a function... so the solution says that I need it in the form $4-(2x+1)^2$, but I cannot figure out how to transform it to that. It could very well be because I'm a bit tired right now, but could somebody point out how it's happening?

$$-4(x+\frac{1}{2})^2 + 4$$ $$-4(x^2 + \frac{1}{2}x + \frac{1}{2}x + \frac{1}{4}) + 4$$ $$-4x^2 - 2x - 2x - 1 + 4$$ $$4 + (-4x^2 - 4x - 1)$$ $$4 - (4x^2 + 4x + 1)$$ $$4 - (2x + 1)^2$$

• I recommend not undoing and then redoing completing the square. Rather, one can use the identity $4(\text{stuff})^2=(2\text{stuff})^2$. – Jonas Meyer Apr 16 '13 at 3:53
• Well I definitely agree, but I thought the OP wanted a very step-by -step answer! – Tyler McAtee Apr 16 '13 at 3:58
• Upvote for your answer! :) – Tyler McAtee Apr 16 '13 at 3:58
• @Mathbro: If you think that $4(\text{stuff})^2=(2\text{stuff})^2$ is not step-by-step, you can just introduce an expression in between: write it as $(4)(\text{stuff}^2)=(2)^2(\text{stuff})^2 = (2\text{stuff})^2$ – ShreevatsaR Apr 16 '13 at 18:58
• You do not need to put it in that form to integrate. The form you have it in is fine.
• You can use the identities $-a+b=b-a$, and $4a^2=(2a)^2$. Note that there is no need to redo your completing the square; you can just absorb the $4=2^2$ into the square.

You can use either forms (they are equivalent) to integrate: the solution you are referring to just offers one form of the function for integrating purposes; it's not the only "correct" form of the function.

Applying the suggestions posted by by @Jonas Meyer:

\begin{align} -4\left(x + \dfrac 12\right)^2 + 4 & \;= \;4 - 4\left( x +\dfrac 12 \right)^2 \\ & =\; 4 - 2^2\left( x +\dfrac 12 \right)^2 \\ & = \;4 - \left(2\left(x + \dfrac 12 \right)\right)^2 \\ \\ & = \;4 - \left(2x+1\right)^2 \end{align}

The process involved in obtaining the given "solution", if you begin from the start, is just another way to "complete the square": \begin{align} 3 - 4x - 4x^2 \; & =\; 4 - 1 - 4x - 4x^2 \\ \\ & =\; 4 - (4x^2 + 4x + 1) \\ \\ & =\; 4 - (2x+1 )^2 \\ \end{align}

giving you a function of the form $a^2 - b^2$ where $a = 2$, and $b = (2x+1)$. Depending on the method used to evaluate the relevant integral, one form may be easier to work with than the other.

• I didn't mention what exactly I was trying to integrate - this is only part of a much bigger question. I needed to convert $\int\frac{6}{\sqrt{3-4x-4x^2}}\ dx$ to a form that I can integrate to $\arcsin{u/a}$. Thanks for the input anyhow. – agent154 Apr 16 '13 at 19:06

$-4(x+ \dfrac{1}{2})^2+4$

$=-2 \cdot2(x^2+\dfrac{1}{4}+x)+4$

$=-4x^2-1-4x+4$

$=4-(4x^2+1+4x)$

$=4-(2x+1)^2$