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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$2012-06-06
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    $Potato I think (s)he wants to factorise stuff here...2012-06-06
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    I am editing the title to change "solving" to "simplifying"2012-06-07
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    Why the downvote?2012-06-07
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    @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**.]2012-06-07

3 Answers 3

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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

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

because

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

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.2012-06-06
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    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.2012-06-06
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    I never took algebra in junior high.2012-06-06
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    ...and then you decide to downvote my answer!?2012-06-06
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    For being rude and talking down to me. Why wouldn't I?2012-06-07
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    Serves me well for being naive...2012-06-07
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    Naive about what? You told me that I was dumber than a junior high kid, that is pretty insulting.2012-06-07
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    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.2012-06-07
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    "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.2012-06-07
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    @AméricoTavares That is fine but talking down to me like this is some basic thing all 12 years olds know and that my having no concept of it is a travesty is taking it too far.2012-06-07
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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.