I Need Math Like a Hobo Needs A Bath
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Author  Topic: I Need Math Like a Hobo Needs A Bath 

Lifecrafter
Member # 34

written Monday, January 8 2007 19:00
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quote:Erm. Well, in fact it is 1.41421356237. Rounded, of course, and presuming the calculator on my iMac and my TI83 Plus are correct. :P Those pictures were from a Facebook group "I Wish I Was A DNA Helicase So I Could Unzip Your Genes." Ironically, the one these were supposedly from had a nerdy pickup line for a title too. [ Monday, January 08, 2007 19:05: Message edited by: Robinator is a Beefcake ]  Frisbeetarianism is the belief that when you die, your soul goes up on the roof and gets stuck. 'Spiderweb Software' anagrammmed: 'Wordbereft A**wipe' Posts: 702  Registered: Wednesday, October 3 2001 07:00 
Off With Their Heads
Member # 4045

written Monday, January 8 2007 19:14
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He was saying that it's not exact, which is obvious. On the SAT, for the purposes of most calculations, the square root of two can be 1.4, and on the MCAT, it can be 1.5 (or more frequently "a little more than 1"). Multiplechoice standardized tests are my life.  Arancaytar: Every time you ask people to compare TM and Kel, you endanger the poor, fluffy kittens. Smoo: Get ready to face the walls! Ephesos: In conclusion, yarr. Kelandon's Pink and Pretty Page!!: the authorized location for all things by me The Archive of all released BoE scenarios ever Posts: 7968  Registered: Saturday, February 28 2004 08:00 
Agent
Member # 5814

written Monday, January 8 2007 19:31
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So I was doing my math homework, and it got really tricky. I'm terrible at math, and since 1) you guys are so clever and 2) this thread's initial reasons for existence are petering out, I decided to request some help. Can someone show (as I already have the answers) me how to do the following problems: 23) sqrt(x)  sqrt(7) = sqrt(x + 7) 25) sqrt(x + 3) + sqrt(x  3) = 3 27) sqrt(2x + 5) + 2sqrt(x + 6) = 5 I know that the general idea is to isolate one radical after another, then square, but I just can't seem to get the numbers to work out. Also, negative square roots don't have to be accounted for (so sqrt(9) is 3 for all intents and purposes, not 3 and 3).  quote: Posts: 1115  Registered: Sunday, May 15 2005 07:00 
Agent
Member # 2820

written Monday, January 8 2007 19:48
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I'm pretty sure that is a sign that you need to move up in course difficulty. Unless you want an easy ride, of course.  Thuryl: I mean, most of us don't go around consuming our own bodily fluids, no matter how delicious they are. ==== Alorael: War and violence would end if we all had each other's babies! ==== Drakefyre: Those are hideous mangos. Posts: 1415  Registered: Thursday, March 27 2003 08:00 
Off With Their Heads
Member # 4045

written Monday, January 8 2007 20:09
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Er, you have an answer for 23? I'd be interested to know which number satisfies that equation.  Arancaytar: Every time you ask people to compare TM and Kel, you endanger the poor, fluffy kittens. Smoo: Get ready to face the walls! Ephesos: In conclusion, yarr. Kelandon's Pink and Pretty Page!!: the authorized location for all things by me The Archive of all released BoE scenarios ever Posts: 7968  Registered: Saturday, February 28 2004 08:00 
Warrior
Member # 7276

written Monday, January 8 2007 20:59
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I find it easier to work with square roots as fractional exponents...that is, instead of sqrt(x), write x^(1/2); instead of sqrt(x + 7), write (x + 7)^(1/2). Your first problem then becomes pretty easy if you square both sides and get your properties of exponents right, and work to get variables on one side and constants on the other. (Though when I actually try to do it, I do get some very nonsensical answers, telling me something is wrong, as per Kelandon. Are you sure you copied that one down right? I am reaching back into the mists of time for this...so I may be forgetting something basic) With the second one, after squaring, I found the problem workable if I moved my nonfractionallyexponented term to the other side and took logarithms of both sides. Remember properties of logarithms  log x + log y = log xy log x^y = y log x and you can raise 10 to a side (antilog) to undo it. This permitted me to reduce the problem to a quadratic equation. (You could get to the same place by isolating your fractionallyexpontented term on one side, then squaring again.) The same technique appears to work for the third one. [ Monday, January 08, 2007 21:32: Message edited by: Alberich ] Posts: 63  Registered: Tuesday, July 4 2006 07:00 
Agent
Member # 5814

written Monday, January 8 2007 21:25
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quote:Oops. I just checked the answer to #23, and it's "No Solution". I'm sorry to have wasted everyone's time on that problem.  quote: Posts: 1115  Registered: Sunday, May 15 2005 07:00 
Lifecrafter
Member # 34

written Tuesday, January 9 2007 13:20
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I kept on reading "sqrt" as "squirt." Things make a lot more sense now, thank you.  Frisbeetarianism is the belief that when you die, your soul goes up on the roof and gets stuck. 'Spiderweb Software' anagrammmed: 'Wordbereft A**wipe' Posts: 702  Registered: Wednesday, October 3 2001 07:00 
Lifecrafter
Member # 6193

written Tuesday, January 9 2007 16:58
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quote:I drove myself crazy trying to do it this way, but to no avail. Just before I wrote myself off as a complete failure, I tried it with just squaring those big nasty numbers and got the right answer. Your way seems like it could be useful, but I just can't get it to work. When you take the log of the left side of the equation you just get a big nasty: Log((x+3)^.5 + (x3)^.5)) Where can I go with this? LogA + LogB = LogAB doesn't work for Log(A+B)  Guaranteed to blow your mind. Frostbite: Get It While It's...... Hot? Posts: 900  Registered: Monday, August 8 2005 07:00 
Warrior
Member # 7276

written Tuesday, January 9 2007 21:05
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I didn't take logs at the beginning, but only after I had squared and grouped terms. This sort of thing  (x+3)^.5 + (x3)^.5 = 3 (squaring both sides): (x+3) + 2(x+3)^.5(x3)^.5 + (x  3) = 9 (grouping terms to get the fractional exponents together) (x+3)^.5(x3)^.5 = 1/2(9  2x) As I mentioned, you could square both sides again; I took logs because I'm a little strange (my grandma always said I liked making life difficult). log(x+3)^.5(x3)^.5 = log 1/2(9  2x) 1/2 log (x^2  9) = log 1/2 (9  2x) log (x^2  9) = log 1/4 (9  2x)^2 As I said, it was really an unnecessary step...but anyway, if you take it from there, it's easy to see it breaks down to a quadratic. The big thing, for me, is working in fractional exponents instead of with radicals. It's conceptually easier and better preparation for calculus. For me, at least, it was easier to remember things like whether you can convert sqrt x / sqrt y into sqrt (x/y) if you wrote them as x^1/2 / y^1/2 . And the properties of logarithms are good things to have in your toolbox, even if they weren't really needed for these problems. [ Tuesday, January 09, 2007 21:12: Message edited by: Alberich ] Posts: 63  Registered: Tuesday, July 4 2006 07:00 
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