Simplifying/Re-arranging Troubles

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The discussion revolves around simplifying a complex statics problem to isolate "a" as a function of other variables. The user struggles with an equation that becomes increasingly messy after attempting to eliminate square roots and rearranging terms. They derive a quadratic equation in terms of "a" but express difficulty in isolating "a" completely. Another participant suggests distributing terms to achieve the standard quadratic form, which allows for the application of the quadratic formula. Ultimately, the goal is to express "a" solely in terms of the other variables involved.
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In the middle of a statics problem full of unknowns, and am trying to simplify a sum of moments down to the point where I have: a= everything else (a as a function of several other terms).

Keep getting close, but then always have an extra variable or two that won't go anywhere.

Here's the equation (the k is just noting a unit vector, so ignore):
http://www.ihostphotos.com/show.php?id=192455"

The first thing I did was multiply through by a factor of sqrt(a^2 + 4r^2) to get rid of that factor in two of the denominators and the sqrt sign on the other, but then it just started getting REALLY messy.

Can anyone figure out what this equation (in the pic) should be, as "a=..." only??

The other variables in that are: P, L, r, and Nb.
 
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Anybody?

It's just algebra, but good lord is it confusing me!
 
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once you get rid of the sq rts, you're left with a quadratic with respect to a. so, group together your a^2, a^1 and a^0 terms, and then plug in the coefficients into the quadratic formula eh?
 
Hmm...after multiplying through by sqrt(a^2 + 4r^2), I got:

[2PLr - Nb(a^2 + 4r^2) + WLa/2 = 0

then I multiplied through by 2 to get rid of the division by 2 in the last term, and got:

[4PLr - 2Nb(a^2 + 4r^2) + WLa = 0

But how do I apply that to the quadratic formula, if I want to get "a" all alone on one side??
 
you can't get a all alone on one side... just like you can't get x on one side if you have x^2 + 2x+3=0. you don't even need to multiply by the two. distribute the N_B and get your equation into ax^2+ bx+c form. then apply the quadratic formula. you'll get two possible answers for a, you'll have to decide which one makes more sense.
 
And that would still end up as "a" as a function of the other terms?

Hmm...going back to my math skills from many years ago...
 
Okay, got it--thanks!
 
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