Trick math question

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thread starter EvilLord30 Apr 16 2010 + PM | QUOTE | PERMALINK | REPORT

If you were smart, you would think this through. Although it may seem easy at first, theres actualy more than one answer! Remember, there are 2 answers, one easier, and the other a bit harder. If your good with math, you'll know this soon enough. Math Question of the day: What is the Square root of "64". Answer is on the bottom of the 1st page of the thread.

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DustInTheWind 04/16/10 + PM | QUOTE | PERMALINK | REPORT

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Enliight 04/16/10 + PM | QUOTE | PERMALINK | REPORT

This kinda fails OT: answers are plus minus 8.

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Supportive 04/16/10 + PM | QUOTE | PERMALINK | REPORT

8^2 Square root sign is ^

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infiniteIQx + PM | QUOTE | PERMALINK | REPORT

[quote] maxxtro : Dude can't you realise that a anything powered by 2 + 1 can not equal 0 unless it's an imaginery number (like i=root of -1)? [/quote] Actually you're right, I just realized what I did wrong... -________________- Solve for all values of x, 0 </= x </= 2pi: 3tan^2(x)-1. MY BAD. Edit: Yeah the other answer was an imaginary number. You spelled imaginary @ person I quoted by the way. Edit 2: The other question was still do-able.

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kole12 + PM | QUOTE | PERMALINK | REPORT

[quote] infiniteIQx : No you can't because you're solving for x, not tan x. I actually didn't give sufficient information. I had incorrect directions... Solve for all values of X in the range 0</= x </= 2pi: 3*tan^2(x)+1=0 Edit: @ above, Please stop typing "homie," it's odd. o_O [/quote] tan^2(x) = -(1/3) the trig identity for tan^2(x) is (1 - cos(2x))/(1 + cos(2x)) (1 - cos(2x))/(1 + cos(2x)) = -(1/3) 1 - cos(2x) = -(1 + cos(2x))/ 3 3 - 3cos(2x) = 1 + cos(2x) (past this point i have no idea whether im going it right) -4cos(2x) = -2 cos(2x) = 1/2 cos(pi/3) = 1/2 so (pi/3)/ 2 = pi/6 which i think is the final answer EDIT: im not gonna do the whole thing again cuz of the sign with the "1" but if you put my answer into the new one it works lmao

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Divination 04/16/10 + PM | QUOTE | PERMALINK | REPORT

does having the square root automatically factor in parentheses or closed-groups? if not, then it's only 8.

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maxxtro 04/16/10 + PM | QUOTE | PERMALINK | REPORT

[quote] infiniteIQx : Actually you're right, I just realized what I did wrong... -________________- Solve for all values of x, 0 </= x </= 2pi: 3tan^2(x)-1. MY BAD. Edit: Yeah the other answer was an imaginary number. You spelled imaginary @ person I quoted by the way. Edit 2: The other question was still do-able. [/quote] Lol if so I guess x=30 degrees or x=pi/6 in radians.

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SaKuRaChiii 04/16/10 + PM | QUOTE | PERMALINK | REPORT

Hmm this might be really simple o_o but... How do you solve x^2 > a^2?

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kole12 04/16/10 + PM | QUOTE | PERMALINK | REPORT

[quote] maxxtro : Lol if so I guess x=30 degrees or x=pi/6 in radians. [/quote] yea you just happened to guess the answer after i went through all the steps....

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DustInTheWind 04/16/10 + PM | QUOTE | PERMALINK | REPORT

[quote] SaKuRaChiii : Hmm this might be really simple o_o but... How do you solve x^2 > a^2? [/quote] sqrt(x^2) > sqrt(a^2) x > a

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Etokapa + PM | QUOTE | PERMALINK | REPORT

[quote] EvilLord30 : Ok I'll answer since I'm getting bored of the thread. 1st answer: 8 2nd Answer: 10! Because 64 = 100 in hex and decimals, and the square root of 100 is 10. [/quote] 10! = 3628800 :P

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SaKuRaChiii 04/16/10 + PM | QUOTE | PERMALINK | REPORT

[quote] DustInTheWind : sqrt(x^2) > sqrt(a^2) x > a [/quote] .. i doubt thats the answer but.. i don't know.

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DustInTheWind + PM | QUOTE | PERMALINK | REPORT

[quote] SaKuRaChiii : .. i doubt thats the answer but.. i don't know. [/quote] That's the answer...I think. Math isn't my forte.

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SaKuRaChiii 04/16/10 + PM | QUOTE | PERMALINK | REPORT

[quote] DustInTheWind : That's the answer...I think. Math isn't my forte. [/quote] Well i already knew that.. but its like.. one step o_o and i don't think it would be that easy

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DustInTheWind 04/16/10 + PM | QUOTE | PERMALINK | REPORT

[quote] SaKuRaChiii : Well i already knew that.. but its like.. one step o_o and i don't think it would be that easy [/quote] That's how you solve those kind of problems though. (x^2)^1/2 > (a^2)^1/2 You're just isolating the x variable. It's right.

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