Not understanding arc midpoint computation

Question

I'm trying to find the midpoint of an arc, so I found this page wherein Gregory V. Akulov and Oleksandr G. Akulov describe the midpoint formula. I pasted the formula & description from the site below.

Let origin-centered arc of radius r have the ends at x=a,b, and midpoint at x=μ. Then

where the first radical gets “−” iff the arc has a negative x-intercept, and the second radical gets “+” iff the arc has a positive x-intercept.

The site poses a example question, also pasted below:

Origin-centered arc of radius 50, located as shown at Figure 2, has the ends at x=14 and x=25. Find x-coordinate of its midpoint.

I tried to solve it, but it seems I went off the rails somewhere.

2μ = ±sqrt((50+14)(50+25)) ±sqrt((50-14)(50-25))
            ↓                          ↓
     ±sqrt(64*75) = ±sqrt(4,800)       ↓
                           ±sqrt(36*25) = ±sqrt(900)
2μ = -sqrt(4,800) +sqrt(900) //x intercept is positive, so both radicals get "+"
μ  = 39.282

But the answer I should have gotten was:

r=50,a=14,b=25, both radicals go with “+”, and 20sqrt(3+15) is the answer;

I haven't taken Trigonometry yet, so maybe that's part of the problem. But this isn't homework, it's something I'm trying to figure out to further a project of my own. There isn't anybody I can ask right now, so I was wondering if someone here could correct my mistake?

It may be helpful to try a simple test case. For example, take the two points to have the same $x$-coordinate but opposite $y$-coordinates. — Semiclassical, Aug 09 '14 at 02:02
If you know both radicals are positive, why did you make the first one negative? — Graham Kemp, Aug 09 '14 at 02:00

Eoin · Accepted Answer · 2014-08-09T02:12:34.580

Your error was just in computation. In your answer you put a negative for the first radical (if it is not a typo).

So we can calculate it again as \begin{gather} \mu =\frac{\sqrt{4800}+ \sqrt{900}}{2} \end{gather} Since $\sqrt{4}= 2$ we can replace it here \begin{gather} \mu = \frac{\sqrt{4800} + \sqrt {900}}{\sqrt{4}} \end{gather} We can seperate the fraction and then join radicals since $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ \begin{gather} \mu = \frac{\sqrt{4800}}{{\sqrt4}} + \frac{\sqrt {900}}{\sqrt{4}} \\\\ \mu = \sqrt{\frac{4800}{4}} + \sqrt{\frac{900}{4}} \end{gather} Now these both divide nicely \begin{gather} \mu = \sqrt{1200} + \sqrt{225} \end{gather} Now the square root of 225 is a nice 15, and 1200 can be written as it's factors $\sqrt{400*3}$ since 400 has a nice root as well. Since $\sqrt{a*b}=\sqrt{a}\sqrt{b}$ we can calculate this as well \begin{gather} \mu = \sqrt{400}\sqrt{3} + 15 \\\\ \mu = 20\sqrt{3} + 15 \end{gather}

Here the radical only extends over the three, it does not include 15. I hope this clears things up for you.

Oh you are right. The… man I must be tired. That being said part of the problem is I'm perhaps a little under-confident in my math skills :. Thanks for the help! — PopKernel, Aug 09 '14 at 02:13
I'm impressed you are doing this by yourself to begin with! Keep it up man, you're doing well. — Eoin, Aug 09 '14 at 02:14

Not understanding arc midpoint computation

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