Doing Parametric differentiation.....finding t problem

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joyoustele
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#1
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How would i get t from this?

-2t+t^2(1+t)^2=0
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RDKGames
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(Original post by joyoustele)
How would i get t from this?

-2t+t^2(1+t)^2=0
Well there is one obvious root, and one not so obvious root.
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joyoustele
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(Original post by RDKGames)
Well there is one obvious root, and one not so obvious root.

t=0

and

t=-2 but i dont know how to get this
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RDKGames
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(Original post by joyoustele)
t=0

and

t=-2 but i dont know how to get this
-2 is clearly not a root. Means your equation is wrong
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Sir Cumference
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(Original post by joyoustele)
t=0

and

t=-2 but i dont know how to get this
Please post the original question. It looks like either you've made a mistake in your working or the equation you posted in your OP is wrong.
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joyoustele
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(Original post by RDKGames)
-2 is clearly not a root. Means your equation is wrong
k, I went wrong some where then, Thanks
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the bear
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factorise:

t { -2 + t ( 1 + t )2 } = 0

then have a think...
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joyoustele
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(Original post by Notnek)
Please post the original question. It looks like either you've made a mistake in your working or the equation you posted in your OP is wrong.
Thanks, worked it out, silly mistake from me
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Sir Cumference
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Trying to solve t(1+t)^2=2 without numerical methods is possible but looks like a nightmare...

Luckily this isn't something that the OP has to do
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Sir Cumference
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(Original post by the bear)
factorise:

t { -2 + t ( 1 + t )2 } = 0

then have a think...
I've had a think, now what
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joyoustele
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(Original post by Notnek)
Trying to solve t(1+t)^2=2 without numerical methods is possible but looks like a nightmare...

Luckily this isn't something that the OP has to do
lol:laugh:
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RDKGames
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(Original post by Notnek)
Trying to solve t(1+t)^2=2 without numerical methods is possible but looks like a nightmare...

Luckily this isn't something that the OP has to do
Just slam the cubic formula on it, what’s so difficult?
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Sir Cumference
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(Original post by RDKGames)
Just slam the cubic formula on it, what’s so difficult?
Yes I suppose you'd need to resort to using the cubic formula here because other methods are too nasty.
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RDKGames
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(Original post by Notnek)
Yes I suppose you'd need to resort to using the cubic formula here because other methods are too nasty.
Just solved it in my head using it.

I’d tell you but this post is too narrow to contain it.
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Sir Cumference
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(Original post by RDKGames)
Just solved it in my head using it.

I’d tell you but this post is too narrow to contain it.
I also did it in my head using a linear adjustment x=t+\frac{2}{3} followed by y=\frac{x}{10} then z=y^3, which turns it into a quadratic.

Simple.
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