Mgccl
here is the image link
OK, this is a bit hard, it took me over 15 minutes to figue out.... SWR's Math 2Honors RULEZ...
Note: ^ means "to the power of"
^1/2 means "squareroot" or "to the power of 1 half"

A math problem only the smart can solveMgccl
here is the image link OK, this is a bit hard, it took me over 15 minutes to figue out.... SWR's Math 2Honors RULEZ... Note: ^ means "to the power of" ^1/2 means "squareroot" or "to the power of 1 half" Jack_Hammer
They are just different ways of writing the same equation, whats so hard about that?
Mgccl
You need to prove that, PROVE why it is equal
cfackler
Warning....this contains the solution. If you want to solve yourself, avoid reading on.
For the repeating radical expression, set it equal to x. Then square both sides of the equation and subtract one from both sides. You can then get x^2  1 = x by replacing the repeating radicals with x. This is the same as x^2  x  1 = 0, solveable with the quadratic formula. With the repeated division...again set the expression equal to x. Then subtract one from both sides. Realize that you have x  1 equals the reciprical of the original x. Thus x  1 = 1 / x. Multiply both sides by x, and you get x^2  x =1, or x ^2  x  1 = 0. Again, solveable with the quadratic formula. (Notice that this is the same equation generated by the repeating radicals.) Plugging 1 for a, 1 for b and c into the quadratic formula... (1 +/ sqrt((1)^2  4 * 1 * 1)) / (2 * 1), which yields (1 + sqrt(5)) / 2. Blaster
wow! That does look hard. I don't get it at all. Then again i am only in Alegabra 1.
bcnhs81
modesty aside, after surviving two terms of calculus and physics, that equation looks kinda simpe enough..
Mgccl
exactly, get (1+square root of 5)/2=a=aa1, then use infinitive substution to get the first forumla, and then make a=sqareroot of(1+a) to solve the second quesion
bassgs_17
Arg... Why take the time to do math anymore, as that's what we have computers for, right!? Just kidding, there would be no computers if it wasn't for that one word... What was it, oh, yes M4TH...
I figured it out with only Precalc Skillz, but it was written rather oddly... Must be my damn US education. Aquastrike Mgccl
go 9th graders.....this was solved by a 9th grader in our school like in.... minutes
hac
Dude I had this question for my test! But instead of all the 1's within square roots we had Six and blah blah...
Mgccl
did this help you in the test
billgertz
wow! that's tough.
theo
not so hard ah?
i think but i didn't even try that um... i think i would like to try later charredii
anyone try it with the squeeze theorem??
i don't have the time or concentration now, but I'm sure it's do able that way sorry, wait till vaca's over! Jack_Hammer
What is so hard about it, and there is no need to prove anything!?, it's like writing,
2 = 1/2 = square root of 4 Just the same thing three times?, why does that need explantation, if I wrote 2 = 2, there is no need to explain it?! Soulfire
I'm in advanced algebra and I can't solve it, but we're half way along. We're currently studying parabolas, converting a standard form quadratic parabola equation to a vertex (and vice versa).
rainynightstarz
oh lookie! its the golden ratio >.< this 1.618 thing
too lazy to prove, had too much series and sequence and limits for the past month... although a nice change from derivatives and intergrals and other junk>.< although i do need practice for AIME... oh well now this is a cool series 1/1+1/2+1/3+1/4......1/n=x as N approches infinity does X converge or diverge? basically asking does X go to a fixed # or infinity? note:1/n as n approches infinity is 0 but you should know that and... Soulfire, u are talking about conics, they are pretty easy as long as you understand how each coeffient affects the graph, and once you learn derivative tests... its even easier, that thats in calculus and i bet a 7th or 8th grader can solve it from my middle school, i mean now i am in highschool ( same district) and taking calc BC with freshmens... cuz my school often scores top on Nationalwide math contest in the US thiamshui
that's too tough.. i hate maths..
Srs2388
i can't complain about using trig in geomtry anymore lol
i understand VERY little of it but some makes a lilttle sense really hard problem cfackler
sum (1/n) with n going from 1 to infinity diverges...simple pseries test.
greenwoodmonkey
Why can women rationally make one add one make three.....
but never figure out how to use the video timer? Bondings
It's not that hard actually, just by defining the whole equation as x and then solve it recursively: x = 1 + 1/(1+...) (the whole thing) Because it's infinite, you can also write it as: x = 1+ 1/x > x^2  x  1 = 0 > x=(1+5^1/2)/2 (must be positive) And for the other one, exactly the same: x=(1+(1+...)^1/2)^1/2 (the whole thing) Because it's infinite, you can rewrite it as: x=(1+x)^1/2 > x^2=1+x > x^2  x  1 = 0 > x=(1+5^1/2)/2 (must be positive) bluespice
This is a totally mind taking question. wish it was easy but im sorry this one got me badly. so what is the final solution, someone care to post???
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