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

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.

-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.

0 is one of the obvious answers, lol

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.

K, lol well i dont have my scientific calculater here with me so give me a bit to figure this one out.

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.

mhm ill keep workin at it when i get my clalc back

thanks. this question is sure interesting.

yeah, it looks like a typical question but its decieving....

the tricky part is 10^x.

 

AND i still can't find who started this thread. Dammn.

yeah really eh. Are you sure this question is possible?

yeah really eh. Are you sure this question is possible?

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.

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.

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...

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...

no idea. i got it from another thread which i could not find. no one seems to be able to find it either.

no idea. i got it from another thread which i could not find. no one seems to be able to find it either.

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.

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.

No really, at least explain to us how to start about doing it, if it's possible without 'Mathmatica'.

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.

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.

(also moved to mathematics forum because this is where it belongs )