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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

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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? :confused: It's a quadratic.

 

Try the quadratic formula. But just by inspection, one answer is 0.

 

The other answer is -2.

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-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.

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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.

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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. :)

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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. :)

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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.

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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.

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