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I am having trouble figuring this out.

$$\sqrt {1+\left(\frac{x}{2}- \frac{1}{2x}\right)^2}$$

I know that $$\left(\frac{x}{2} - \frac{1}{2x}\right)^2=\frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}$$ but I have no idea how to factor this since I have two x terms with vastly different degrees, 2 and -2.

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I am trying to find the arc length of $y = 1/4x^2 - 1/2 lnx$ – user138246 Jun 6 '12 at 23:49
$Potato I think (s)he wants to factorise stuff here... – DonAntonio Jun 6 '12 at 23:51
I am editing the title to change "solving" to "simplifying" – Américo Tavares Jun 7 '12 at 0:15
Why the downvote? – Argon Jun 7 '12 at 0:43
@Argon Make some browsing in his profile - specially, see the number of questions asked, the number of questions that are essentially the same problem with different variables, and how the OP doesn't put much of his effort in either solving the problem or understanding the hints or help of other users. [No offense intended to anyone.] – Pedro Tamaroff Jun 7 '12 at 0:49
up vote 6 down vote accepted

Since $$\begin{equation*} \left( \frac{x}{2}-\frac{1}{2x}\right) ^{2}=\frac{x^{2}}{4}-\frac{1}{2}+ \frac{1}{4x^{2}}, \end{equation*}$$

we have

$$\begin{eqnarray*} 1+\left( \frac{x}{2}-\frac{1}{2x}\right) ^{2} &=&1+\left( \frac{x^{2}}{4}- \frac{1}{2}+\frac{1}{4x^{2}}\right) \\ &=&1+\frac{x^{2}}{4}-\frac{1}{2}+\frac{1}{4x^{2}} \\ &=&\frac{x^{2}}{4}+\left( 1-\frac{1}{2}\right) +\frac{1}{4x^{2}} \\ &=&\frac{x^{2}}{4}+\frac{1}{2}+\frac{1}{4x^{2}} \\ &=&\left( \frac{x}{2}+\frac{1}{2x}\right) ^{2}, \end{eqnarray*}$$


$$\left( \frac{x}{2}+\frac{1}{2x}\right) ^{2}=\frac{x^{2}}{4}+\frac{1}{2}+ \frac{1}{4x^{2}}.$$

Therefore $$\begin{equation*} \sqrt{1+\left(\frac{x}{2}-\frac{1}{2x}\right)^2}=\sqrt{\left(\frac{x}{2} +\frac{1}{2x}\right)^2}=\left\vert\frac{x}{2}+\frac{1}{2x}\right\vert . \end{equation*}$$

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Hint: for any (real, complex) numbers $\,a,b,\,$: $$4ab+(a-b)^2=(a+b)^2$$

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So this is a form I need to have memorized? I also do not understand how that applies to this problem. – user138246 Jun 6 '12 at 23:49
Well, these are Junior High School algebra formulae which are so in use in every day mathematics that at some point, I guess, people get used to them as much as to the multiplication tables. – DonAntonio Jun 6 '12 at 23:53
I never took algebra in junior high. – user138246 Jun 6 '12 at 23:56
No, I didn't. I specified to you that these formulae are j.h.s. stuff and so widely used that people usually gets used to them, and this as an answer to your odd comment "So this is a form I need to have memorized? I also do not understand how that applies to this problem", which is not only whinning but also shows you've not made any effort to understand the hint. I guess that after almost 200 questions you've asked (more than 30 just in the last week), you're already used to people doing your homework for you. – DonAntonio Jun 7 '12 at 0:14
"I never took algebra in junior high" @Jordan, When one wants to "find the arc length" of a curve given by an elementary function, one normally knows some elementary algebra. – Américo Tavares Jun 7 '12 at 1:26

I presume you aren't asked to solve this (since it isn't an equation), but rather are asked to express it in a tidier form. Carrying on, we have \begin{align*} 1+\left(\frac{x^2}{4}-\frac{1}{2}+\frac{1}{4x^2}\right) &=\frac{x^2}{4}+\frac{1}{2}+\frac{1}{4x^2}\\\\ &= \left(\frac{x}{2}+\frac{1}{2x}\right)^2\\\\ &= \left(\frac{x^2+1}{2x}\right)^2 \end{align*}

Carry on from there: put the whole expression under the radical, use the fact that $\sqrt{a^2}=\mid a\mid$ to get $$ \left|\frac{x^2+1}{2x}\right| $$

By the way, this idiom, $4ab+(a-b)^2=(a+b)^2$, is very common and should eventually be part of your mathematical toolkit.

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Regarding your response to DonAntonio's comment, a very important part of solving math problems is the ability to recognize patterns. He and I were trying to point out a useful pattern, where one recognizes that the expressions (a + b)^2 and (a- b)^2 only differ by 4 ab. The situation in this problem is just that, for suitable substitutions of a and b. The ability to abstract, to recognize a general pattern as one you've seen, will serve you far better than memorizing formulas. Good luck. – Rick Decker Jun 7 '12 at 0:53
Continuing my comment, you've posted two arc length questions, the solutions to both of which have involved the same idea, that 1 - <some square> = <another square>, so when you see a similar problem, perhaps on a homework or exam, your first thought should be "can I use the technique I've seen on this new arc length problem?" For arc length problems in particular, generally the only exercise-suitable problems (i.e., those that students can actually do) will probably involve this idiom. – Rick Decker Jun 7 '12 at 1:11

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