gene Posted November 6, 2004 Posted November 6, 2004 Spent some time finding the thread that asked how to find 20x + 10x^2 = 0 So, moderators, please help. Sorry. Ok, you can solve graphically although i know the answer might not be accurate. By trial and error, i got an answer where x=0.03... something. I guess you can't solve by log as you can't log a negative 20x. the answer should be from between y=-20x and y=10x^2 I think only by using a special software to graphically draw out the graph and read from it. Hope it might help. :x
swansont Posted November 6, 2004 Posted November 6, 2004 Spent some time finding the thread that asked how to find 20x + 10x^2 = 0 So' date=' moderators, please help. Sorry. Ok, you can solve graphically although i know the answer might not be accurate. By trial and error, i got an answer where x=0.03... something. I guess you can't solve by log as you can't log a negative 20x. the answer should be from between y=-20x and y=10x^2 I think only by using a special software to graphically draw out the graph and read from it. Hope it might help. :x[/quote'] special software? log? It's a quadratic. Try the quadratic formula. But just by inspection, one answer is 0. The other answer is -2.
Fellowes Posted November 7, 2004 Posted November 7, 2004 -2 is what i got just doing it in my head now......... x=(-2) 20x + 10x^2= 0 20 times -2 = -40 -2^2 = 4 times 10 = +40 -40 + 40 = 0 I'm only in junior high school and unless this problem is something totally different than that, then those are my thoughts on that question.
gene Posted November 10, 2004 Author Posted November 10, 2004 Shoots!!! The equation is suppose to be 20x+10^x = 0 Sorry folks. I remembered wrongly... anyone know's where is the thread??? can't find.
Fellowes Posted November 10, 2004 Posted November 10, 2004 K, lol well i dont have my scientific calculater here with me so give me a bit to figure this one out.
gene Posted November 11, 2004 Author Posted November 11, 2004 yeah. can't seem to get the answer. only be computer's plotting of graph then i think an answer would be possible. i ask my teacher on that one. he talked about searching it through graphs. log would be impossible. No reply since.
Fellowes Posted November 11, 2004 Posted November 11, 2004 mhm ill keep workin at it when i get my clalc back
gene Posted November 12, 2004 Author Posted November 12, 2004 thanks. this question is sure interesting.
Fellowes Posted November 14, 2004 Posted November 14, 2004 yeah, it looks like a typical question but its decieving....
gene Posted November 14, 2004 Author Posted November 14, 2004 the tricky part is 10^x. AND i still can't find who started this thread. Dammn.
Fellowes Posted November 19, 2004 Posted November 19, 2004 yeah really eh. Are you sure this question is possible?
Fellowes Posted November 19, 2004 Posted November 19, 2004 yeah really eh. Are you sure this question is possible?
gene Posted November 19, 2004 Author Posted November 19, 2004 not sure. but graphically it is possible. if you have that so called "advance mathematics software plotting system" ha. forget it, i can't get it too. Althought it is food for thought! Thanks for your help.
gene Posted November 19, 2004 Author Posted November 19, 2004 not sure. but graphically it is possible. if you have that so called "advance mathematics software plotting system" ha. forget it, i can't get it too. Althought it is food for thought! Thanks for your help.
Xandrabeast Posted November 19, 2004 Posted November 19, 2004 Doesn't x have to be 0? I can't see it working any other way...oh wait, nevermind...it can't be zero. ...by the way, where was this equation found? I mean, besides the thread that can't be found...
Xandrabeast Posted November 19, 2004 Posted November 19, 2004 Doesn't x have to be 0? I can't see it working any other way...oh wait, nevermind...it can't be zero. ...by the way, where was this equation found? I mean, besides the thread that can't be found...
gene Posted November 19, 2004 Author Posted November 19, 2004 no idea. i got it from another thread which i could not find. no one seems to be able to find it either.
gene Posted November 19, 2004 Author Posted November 19, 2004 no idea. i got it from another thread which i could not find. no one seems to be able to find it either.
Joshua Posted November 19, 2004 Posted November 19, 2004 OK. I'm on the case. Let's crack this baby open! A little experimentation shows the answer must be very close to zero and it must be negative. I could post the answer now(I used Mathematica) but that'd be no fun! Let's figure out how to do it algebraicly. Hint: I don't know the exact procedure but this CAN be done with logs.
Joshua Posted November 19, 2004 Posted November 19, 2004 OK. I'm on the case. Let's crack this baby open! A little experimentation shows the answer must be very close to zero and it must be negative. I could post the answer now(I used Mathematica) but that'd be no fun! Let's figure out how to do it algebraicly. Hint: I don't know the exact procedure but this CAN be done with logs.
Fellowes Posted November 23, 2004 Posted November 23, 2004 No really, at least explain to us how to start about doing it, if it's possible without 'Mathmatica'.
gene Posted November 23, 2004 Author Posted November 23, 2004 Josh. i disagree. It CAn't be done with log because we can't log a negative number. 10^x + 20x = 0 10^x = -20x log 10^x = log (-20x) ---> Impossible.
Dave Posted November 23, 2004 Posted November 23, 2004 That statement isn't impossible, it just implies x is negative. Also, I'm not sure whether you can solve the equation analytically - looks like a transdental equation to me.
Dave Posted November 23, 2004 Posted November 23, 2004 (also moved to mathematics forum because this is where it belongs )
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