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sqrt(2005*2003*1999*1997+36)

Will you be able to solve this without using a calculator?

4003991 (with calculator)
Does the answer need to be exact or approximate? You can find something strikingly similar to the above answer by averaging the number multiplying (2001), multiplying it by itself, and adding the square root of 36 (6). Not sure if this is universal, but I think it works. Using this method, I get 4004007. 18 away from the needed answer.

No way.

2001 is important here. +1 ego boost to Centreri.

The one that posts the way of simplyfying this will get a +1337 ego boost.

So it has to be exact? Time to start calculatorin' me ma' own theorem!

Yes, it must be exact.

ADDITION:
Common guys, don't let DTBK get all the ego boost!

I tried. I don't noez teh rite theoremz. I'll tell you when I have a brainstorm . I keep getting close, but always 5-25 wholes off .

Edit: Damn! I got is just four wholes off. Stupid math. Still wrong.

What theorems do you need to know to solve this? I think you got all the knoweledge needed in about 7th or 8th grade.

Nah, I never did stuff like this in seventh or eighth grade. There's probably a theorem specifically for the 'four numbers centered around one multiplied' scenario.

I don't know such a theorem. You don't need it anyway.

I am not doing the Babylonian or Newton's methods on such a huge number at 12. I have an philosophy; don't try to solve square roots manually when calculators can do it for you.

I wish teachers saw it that way. But nooo, computers and calculators are evil.

You must first make it simpler, and bring it to a point where one multiplication action will solve it. (and one more simple action) And no, you won't have to multiply huge numbers.

ADDITION:

They are to an extent. They do the thnking you can do yourself for you.

Fine, try to make Einstein find the square root of 1129312341096234^123471234.

My math teacher saw it that way.

6 sqrt(4016015*3992003)

6 sqrt(16031943928045)

6*4003990.9¯

24023945.9¯

ADDITION:


1129312341096234[sup]6173561.5[/sup]

How is that on-topic? Einstein can't calculate that. I was just saying that people use machines to think for themselves, where they could have done it without the help of the machines with little loss of time.

Variables are so much easier... still working on it, not finished yet.
square root((x-4 . x-2 . x+2 . x+4) + y)
x^2 - 6x + 8
x^3 - 6x^2 + 8x + 2x^2 - 12x + 16
x^3 - 4x^2 - 4x + 16
x^4 - 4x^3 - 4x^2 + 16x + 4x^3 - 16x^2 - 16x + 64
x^4 -20x^2 + 64
x^4 - 20x^2 + 100 & x=2001
Bah.. I forgot how to take square roots of unsolved equations...

You stopped before the easy part.

ADDITION:
Your edit is incorrect.

I'm supposed to do this without the aid of the calculator. So I have to find the square root before plugging in the 2001. Which I forgot how to do.

Anyone want to take over? I take half of the 1337 .

Why the hell did you do it the hard way? I'm sure you know formulas like this: a[sup]2[/sup]+b[sup]2[/sup]=(a+b)(a-b)

In a world where we go into the fourth power and multiply by four equations, I want to do it the hard way to make sure. Stop making me feel bad. I'm stumped.

You have:
(x-4)(x-2)(x+2)(x+4)+36=(x[sup]2[/sup]-16)(x[sup]2[/sup]-4)+36=x[sup]4[/sup]-20x[sup]2[/sup]+100

Square root of that. And my brain got stupid and forgot how to find the square root of that. Without using a calculator.

x[sup]4[/sup]-20x[sup]2[/sup]+100=(x[sup]2[/sup]-10)[sup]2[/sup]

There you have it. Multiply 2001 by 2001 and subtract 10.

I missed that? Ouch... Do I get half of 1337 for almost making it?