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April 18th, 2010, 03:48 AM   #1
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can this be simplified for S?

I was thinking if the formula below can be solved for S?

L = 8Dsquared/3S + S

I already got the answer through the quadratic as numbers were given for L (80) and D (2.5). But just for the heck of it I tried to resolve it for S. Multiple S variables keep fouling me up no matter what route I have tried.

E.g., moving S by itself but then one has to engage the 3S somehow. L - 8Dsq./3S = S. Even factoring out did not seem to help. I suspect I'm missing the simplicity of the simplification! :- D

Perhaps L (3S) - 8Dsq. = 3Ssquared is a start?

And then...

L (3S) - L(3S) - 8Dsq. = 3Ssq. - L(3S) giving -8Dsq. = 3Ssq. - L(3S) ?

But even with factoring I'm not sure how to simplify these multiple S variables. -8Dsq. = 3S (S - L).

This is just for fun so no rush. I am also wondering if there just are equations which cannot be simplified? Seems that they all should be subject to simplification, however.
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April 18th, 2010, 04:26 AM   #2
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Did you mean L = 8D/(3S) + S or L = (8D/3)S + S?

The first of the above alternatives leads to 3S - 3LS + 8D = 0, and you can solve that by using the quadratic formula.
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April 18th, 2010, 03:39 PM   #3
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Re: can this be simplified for S?

Oh, no problem. As noted, I got the answer by plugging in the numbers and using the quadratic etc. as in 3Ssq. - 240S = -50.

What I was wondering was whether the form could be simplified to S without the numerals; i.e., S = etc. Can this equation be simplified to S without putting in the numbers?
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April 18th, 2010, 03:56 PM   #4
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You get S = (3L ?(9L - 96D))/6.
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April 18th, 2010, 06:50 PM   #5
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Re: can this be simplified for S?

OK, this is great, but just to confirm what I meant is can this equation be simplified without use of the numbers? I.e., does the form you presented here use only the coefficients given in the original form (i.e., 8Dsq. and 3S)?

If we did not know that D = 2.5 and L = 80 (or whatever the numerical equivalents were) could one simplify for S given the original form of L = 8Dsq./3S + S ?

Just wondering if the form here was found without the D and L numerical equivalents, that's all. If so, good show! And...could you show the steps? I think once we get into the squaring and the square roots that's where I foul up.... Danke!
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April 19th, 2010, 06:10 AM   #6
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What I gave was based on the quadratic formula. It works for any values of D and L.
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April 21st, 2010, 04:28 PM   #7
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Re: can this be simplified for S?

OK, that's good. Just wanted to know if it could be done without the use of the quadratic. So without the quadratic, it cannot be simplified in other ways (e.g., by factoring out this or that, multiplying by this or that, etc.).

Otherwise the equation was driving me crazy! :- /
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April 21st, 2010, 05:05 PM   #8
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Re: can this be simplified for S?

If you can use the quadratic equation, it's just as good as any other method of factoring the polynomial into (x - root1)(x - root2) = 0. If you really feel guilty using it, you could first prove the quadratic equation constructively, and then use it to solve for S.
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April 21st, 2010, 06:33 PM   #9
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Re: can this be simplified for S?

All right... sounds good... Thanks yet again!

Thus, for the original form of L = 8Dsquared/3S + S, and without using the numerical values given in the problem (but using the coefficients), we can still use the quadratic. That's fine.

Now I'm just trying to figure how skipjack obtained the form before he used the quadratic.

For instance, so far the best I can do (which is not so good!) is the following:

1. S = L - 8Dsq./3S

2. 3Ssq. = 3SL - 8Dsq.

3. 3Ssq. - 3SL + 8Dsq. = 0

So I'm assuming IF this is correct, that my quadratic numbers for a, b, and c are 3, -3, and 8? But does that make sense given that D is squared?

Sorry to drag this out... Very curious about this....
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April 21st, 2010, 09:23 PM   #10
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By comparing 3S - 3LS + 8D and ax + bx + c, a = 3, b = -3L, and c = 8D.
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