Yes, you are correct. Your answer also evaluates to 2.26, which is a nice check. Algebra is a bitch.
None.
It's called the Quadratic Formula in English (x = (-b +/- sqrt(b2 - 4ac)) / 2a).
Yes, you are correct. Your answer also evaluates to 2.26, which is a nice check. Algebra is a bitch.
Yes, you are correct. Your answer also evaluates to 2.26, which is a nice check. Algebra is a bitch.
yeah.. I am not seeing how to work it out with the quadratic formula. Can you elaborate more?
I am a Mathematician Computer Scientist Logic user
>be faceless void >mfw I have no face
Quadratic equation works once you have your equation in the form ax^2+bx+c. So for this problem, a = 2^x, and b = -5, and c = 1. Then you put those into the quadratic equation and the anrwer is what x equals.
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
"In short, their absurdities are so extreme that it is painful even to quote them."
Oh god, a solve my homework topic...
Anyway...
When solving an exponential equation the simplest way is to get all the powers to look the same. In this case you got 2^x and 2^(-x) which can be easily converted to 1/2^x:
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
Anyway...
Quote
2^x + 2^(-x) = 5
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
?????
Actually in this case, a was 1 and x itself was 2^x.
None.
Didn't there used to be a rule against posting topics for people to do your homework?
None.
Dungeon Master
No, homework solving threads have been quite common over the past few iterations of SEN.
Quote from 
Clokr_
Clokr_Oh god, a solve my homework topic...
Anyway...
When solving an exponential equation the simplest way is to get all the powers to look the same. In this case you got 2^x and 2^(-x) which can be easily converted to 1/2^x:
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
Anyway...
Quote
2^x + 2^(-x) = 5
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
I remember browsing through some topics in serious discussion, I saw a couple of calculations at your hands. I was hoping you would come along!
I'm only here because they patched SC1 and made it free.
Quote from 
Dapperdan
DapperdanDidn't there used to be a rule against posting topics for people to do your homework?
I wouldn't mind helping to solve homework as long as they ask how and not what is the answer.
Still SEN might not be the right place to ask this kind of questions.
Quote from ProtoTank
I remember browsing through some topics in serious discussion, I saw a couple of calculations at your hands. I was hoping you would come along!
That's what we math majors are for
?????
Look on first page >_< I want some credits too.
*rawrs
*rawrs
None.
You did help as well, you did basically the same thing as clokr_, he just substituted. His method just clicked a little easier. Thanks for the contribution anyway!
I'm only here because they patched SC1 and made it free.
I always thought factoring was 10x easier than the quadratic
None.
Master has given Dobby a doctorate! Dobby is free!
Quote from name:Raccoon
I always thought factoring was 10x easier than the quadratic
I always thought the quadratic worked 10x as often as factoring
BeDazed solved it essentially the same as Clokr, so give him some credit
None.
Quote from poison_us
poison_usQuote from name:Raccoon
I always thought factoring was 10x easier than the quadratic
I always thought the quadratic worked 10x as often as factoring
None.
Quote
your answer is wrong 
. I respect you as a forum member, but that is not the correct way to solve the problem graphically.
. I respect you as a forum member, but that is not the correct way to solve the problem graphically. Woops, guess I plugged it into WinCalc wrong
(although admittedly I should've been able to easily do that in my head..) All this logarithm and fractional exponents and such were done in 5/6 math last year, but I forgot it Although I still don't understand why my way of solving doesn't work properly, it seems like it should get the answer..
None.
Quote from name:Ultimo
Quote
log2(5-2x/2x) = -2x
Quote
-(5-2x)log2(2x) = -2x
I figured log(1/n) = -log(n), so why not?
None.
Quote from Clokr_
Clokr_Oh god, a solve my homework topic...
Anyway...
When solving an exponential equation the simplest way is to get all the powers to look the same. In this case you got 2^x and 2^(-x) which can be easily converted to 1/2^x:
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
Anyway...
Quote
2^x + 2^(-x) = 5
2^x + 1/2^x = 5
Now you do a = 2^x to get a standard equation:
a + 1/a = 5
You solve it for a:
(a²+1)/a = 5 -> a²+1 = 5a -> a² - 5a + 1 = 0
so a = (5ħsqrt(25-4))/2 = (5ħsqrt(21))/2
Since a = 2^x then 2^x = (5ħsqrt(21))/2 and thus x = log2 ((5ħsqrt(21))/2)
Solved
I cannot believe I didn't think of solving via substitution.
None.
Quote
I figured log(1/n) = -log(n), so why not?
log(ax) = log(a) + log(x), not a*log(x).
None.
Master has given Dobby a doctorate! Dobby is free!
Quote from name:Raccoon
Quote from poison_us
poison_usQuote from name:Raccoon
I always thought factoring was 10x easier than the quadratic
I always thought the quadratic worked 10x as often as factoring
You have me curious: what's x² + 21x + 85 = 0?
[2026-7-18. : 2:27 am]
Symmetry -- https://staredit.net/topic/18903/ THE REVIEWS ARE HERE! "ok these are pretty good memes" - ssj9Kevin[2026-7-17. : 2:58 pm]
Ultraviolet -- SymmetrySymmetry shouted: Man I wish we had this tech back in the day when people played ums lmao
fr fr[2026-7-17. : 2:39 am]
Symmetry -- Man I wish we had this tech back in the day when people played ums lmao[2026-7-16. : 6:28 am]
NudeRaider -- EUD actions, specifically. The conditions remained. Today we can even have actions again, but they're all virtualized/whitelisted (not sure on the actual technical implementation) so you can only use them for specific intended purposed, not arbitrary code injections.[2026-7-16. : 6:26 am]
NudeRaider -- I mean yes, maps - through EUDs - were theoretically able to do that as well, but that was patched quickly.[2026-7-16. : 6:25 am]
NudeRaider -- SymmetrySymmetry shouted: Ohhh imagine a SC map that could delete itself
not the map, the editor[2026-7-16. : 2:28 am]
Symmetry -- I vaguely remember Voy doing some ACE back in the day, but I had no idea EUDs could do that kind of shit. I am very much catching up on the technology[2026-7-15. : 8:51 pm]
NudeRaider -- SymmetrySymmetry shouted: NudeRaider Is EUD editor capable of writing shit onto the player's harddrive? That seems like it would be dangerous
yeah an editor that isn't allowed to write files sounds rather pointless, so I'd assume it can. Like most other programs too btw. and yes, obviously that's dangerous, but also kinda necessary. MS introduced that virtualization for exactly that reason.[2026-7-15. : 4:10 am]
Ultraviolet -- I don't think so. I think they shut that shit down after 1.16, maybe even earlier
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