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**UrgentHelp****Guest**

Hi, I have spent SOOO long trying to manipulate this thing to get the right answer.. I really cannot do it.

Could you please help me? We have:

1/a = (2(1-u))/(((1-u)^2)-4g) + 2/(1+u)

**UrgentHelp****Guest**

The Answer is:

a= (SQUAREROOT of 5-4b) - 2.

Please please help me! I really can not manipulate it to get this solution

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,247

Hi;

Where does the b come from?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi Urgenthelp,

And where did u and g go?

Is this your expression?

Bob

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**UrgentHelp****Guest**

Yes bob bundy, that is exactly it!

Bobbym, sorry that should =g, not b, really sorry.

I am getting so many notes but i cannot get that solution for u=...

**UrgentHelp****Guest**

So to clarify the eventual solution is supposedly:

(SQUAREROOT of (5-4g)) - 2

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi Urgenthelp,

So we are getting closer. But I still don't see where a has gone. Is there another equation? Where did this equation come from?

Bob

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

Another thought. Could that 'a' be a nine ?

Bob

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**UrgentHelp****Guest**

bob bundy wrote:

Another thought. Could that 'a' be a nine ?

Bob

I AM SO SORRY. I do not know why I cannot type, that a is meant to be 'u' just like the 'u' on the RHS.

The whole thing comes from differentiating, so u=(squareroot 5-4g) - 2

is the maximum for whatever g is. I am SO sorry for giving that typo and wasting your time. It is meant to be 1/u on the LHS too

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi Urgenthelp

Got it. Typing it up now. I'll post every so often and then edit the post to add the latest bit. That way I won't lose any of the bits (I hope)

multiply all terms by the common denominator to clear all fractions

collect all terms on the LHS

That's looks pretty nasty but I notice that putting u = 1 makes this expression zero => (u-1) is a factor

So one solution is u = 1 and I'll use the quadratic formula for the other two

The plus sign case is the one you want. Presumably you can find a reason to disregard the other two.

Bob

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**UrgentHelp****Guest**

This is the best post I have EVER seen. Thank you SOOO MUCH. You are incredible. I really don't know how you thought to take that route, it is brilliant. Really thank you!

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi Urgenthelp,

Thanks for the nice reply!

I always try to avoid fractions in algebra, so the first step was to get rid of them. Then I hoped that lots of factors would cancel, but no such luck. So there was nothing to do but 'wade through' all that horrible algebra, hoping to avoid any slip ups.

When I got to the cubic I thought, "Oh no, that looks impossible" But the answer you had posted suggested that the quadratic formula should be used. So how to get to a quadratic? Only by extracting an easy factor first. .... so I looked for one. Tried u = 1 first (because it's the easiest) and it worked! That encouraged me to think I was on the right lines and that I hadn't slipped up with the algebra so far.

The rest was routine, find the quadratic by examination, apply the formula, simplify. Observe one answer is what you wanted. Feel smug!

Bob

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