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Author Topic: Math!  (Read 3103 times)

Nick1911

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Math!
« on: January 30, 2011, 06:53:21 PM »
How would you solve this?



My inclination is to square all terms, but this will strip the negative value out, and leave me with no real solutions.

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vaskidmark

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Re: Math!
« Reply #1 on: January 30, 2011, 07:02:30 PM »
Quote
author=Nick1911 link=topic=28085.msg549386#msg549386 date=1296431601]
How would you solve this?

"YES"  if x = 4.

Too lazy to see if substituting any other number for the (-n), (sq root of x) and <(n) values will also work.

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Perd Hapley

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Re: Math!
« Reply #2 on: January 30, 2011, 07:03:31 PM »
0 < x < 16 ?
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RoadKingLarry

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Re: Math!
« Reply #3 on: January 30, 2011, 07:11:48 PM »
Bacon
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Jim147

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Re: Math!
« Reply #4 on: January 30, 2011, 07:24:38 PM »
Larry wins!

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Re: Math!
« Reply #5 on: January 30, 2011, 08:14:10 PM »
yep the answer is 4

-4 is less then 2 is less then 4

2 being the square root of 4.

You're over thinking the problem. 
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Re: Math!
« Reply #6 on: January 30, 2011, 08:30:39 PM »
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How would you solve this

Let my TI-89 do it. Let's put it this way...I can barely do very basic math. That calculator saved me in school...

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Re: Math!
« Reply #7 on: January 30, 2011, 08:35:23 PM »
couldn't the answer be up to 16?  also, how can you get the square root of a negative number?
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Re: Math!
« Reply #8 on: January 30, 2011, 08:39:52 PM »
It could be up to 9, if it were 16 then it would be -4 < 4 < 4
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zahc

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Re: Math!
« Reply #9 on: January 30, 2011, 08:44:25 PM »
Quote
yep the answer is 4

There is a whole set of answers, not just one.

Quote
It could be up to 9, if it were 16 then it would be -4 < 4 < 4

How do you go from 16 to 9? why not 15.99? Did someone specify that x was an integer?

If x must be real, then the answer is [0,16).



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Re: Math!
« Reply #10 on: January 30, 2011, 08:51:26 PM »
I assumed the integer part.  I suppose if that assumption were not made then yes 15.99 and on down would work.
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Nick1911

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Re: Math!
« Reply #11 on: January 30, 2011, 08:58:29 PM »
Quote from: zahc on January 30, 2011, 08:44:25 PM
There is a whole set of answers, not just one.

How do you go from 16 to 9? why not 15.99? Did someone specify that x was an integer?

If x must be real, then the answer is [0,16).

Correct.

Okay, so here's how I ended up solving it.

Break into two equations

1.
sqrt(x) < 4
x < 16

sqrt(x) > -4
No negative numbers can make this equation true.  The smallest number for which this holds true is 0.

Therefore, [0, 16)

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Re: Math!
« Reply #12 on: January 30, 2011, 09:10:23 PM »
Forgive me..  it has been a lot of years since I took a math class. 

But wouldn't [0,16] infer that 16 is one of the possible answers? 
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Re: Math!
« Reply #13 on: January 30, 2011, 09:13:17 PM »
Fistful's got it.

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Nick1911

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Re: Math!
« Reply #14 on: January 30, 2011, 09:15:28 PM »
That's why we used the open interval on the end.  http://mathworld.wolfram.com/OpenInterval.html

The answer includes 0, but not 16.  Everything up to 16, though.

Quote from: Doggy Daddy on January 30, 2011, 09:13:17 PM
Fistful's got it.

DD

Nope.  Fistfuls answer doesn't include 0, and 0 is one of the answers.

Thanks for the help, all.  I think I got it figured out.

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Re: Math!
« Reply #15 on: January 30, 2011, 09:25:36 PM »
Quote from: Nick1911
Nope.  Fistfuls answer doesn't include 0, and 0 is one of the answers.

Correct you are.  And we see once again that Fistful IS the master of deception.

DD
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Re: Math!
« Reply #16 on: January 30, 2011, 09:27:41 PM »
oh yeah!  I remember that now!
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Tallpine

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Re: Math!
« Reply #17 on: January 31, 2011, 01:56:26 PM »
What about -16i < x < 16  ???

i = square root of -1 (imaginary number)
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Perd Hapley

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Re: Math!
« Reply #18 on: January 31, 2011, 02:05:55 PM »
The square root of -16i would be (4i)(square root of i). Right?

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Re: Math!
« Reply #19 on: January 31, 2011, 02:06:26 PM »

i sqrt(bacon)

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Re: Math!
« Reply #20 on: January 31, 2011, 02:17:00 PM »
Quote from: Fistful on January 31, 2011, 02:05:55 PM
The square root of -16i would be (4i)(square root of i). Right?



I imagine so...
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Perd Hapley

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Re: Math!
« Reply #21 on: January 31, 2011, 02:54:48 PM »
So why do you want that extra i there in the first place?
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Tallpine

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Re: Math!
« Reply #22 on: January 31, 2011, 03:02:26 PM »
Quote from: Fistful on January 31, 2011, 02:54:48 PM
So why do you want that extra i there in the first place?

Just trying to figure out if there are any solutions <0  ???

I imagine that any solutions <0 would be imaginary  :lol:
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Re: Math!
« Reply #23 on: January 31, 2011, 07:24:11 PM »
Seems to me that was soluble by inspection.

Isn't that a valid arithmetical operation/method?

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