Can this equation be solved for F?
Can this equation be solved for F?
F/3500 = [85+(1750/F)^2)^-.5] + [74+(1750/F)^2)^-.5]
Just to be clear, this equation is read:
(F divided by 3500) equals (85 plus (1750 divided by F) squared) all to the negative one half plus (74 plus (1750 divided by F) squared) all to the negative one half.
Sorry I don't know latex.
...And for your own sake do not try to solve this equation. I am 99% sure it cannot be done unless it involves a computer program and/or imaginary numbers.
I am just hoping someone can tell me if it can or cannot be solved and why.
Re: Can this equation be solved for F?
Re: Can this equation be solved for F?
Quote:
Originally Posted by
piscoau 
Thanks for the reply but you read the equation incorrectly. The 85 and the 74 are part of the quantity that is raised to the power of -.5
I wish it were as simple as you put it...but it is not.
Anyone know?
Re: Can this equation be solved for F?
If I'm correct, the equation is:
![\frac{F}{3500}=\left[85+\left(\frac{1750}{F}\right)^2\right]^{-0,5}+\left[74+\left(\frac{1750}{F}\right)^2\right]^{-0.5}](http://latex.codecogs.com/png.latex?\frac{F}{3500}=\left[85+\left(\frac{1750}{F}\right)^2\right]^{-0,5}+\left[74+\left(\frac{1750}{F}\right)^2\right]^{-0.5})
Because it's an irrational equation, have you tried to square both sides?
Re: Can this equation be solved for F?
Noting that 3500= 2(1750), I would first simplify the notation by letting 
so that the equation becomes
![\frac{x}{2}= \left[85+ \frac{1}{x^2}\right]^{-0.5}+ \left[75+ \frac{1}{x^2}\right]^{-0.5}](http://latex.codecogs.com/png.latex?\frac{x}{2}= \left[85+ \frac{1}{x^2}\right]^{-0.5}+ \left[75+ \frac{1}{x^2}\right]^{-0.5})
.
And, to get rid of the negative exponents, I would multiply both sides by the bracketed quantities: if 
, then 
. Here, that gives ![\frac{x}{2}\left[85+ \frac{1}{x^2}\right]^{0.5}\left[75+ \frac{1}{x^2}\right]^{0.5}= \left[85+ \frac{1}{x^2}\right]^{0.5}+ \left[75+ \frac{1}{x^2}\right]^{0.5}](http://latex.codecogs.com/png.latex?\frac{x}{2}\left[85+ \frac{1}{x^2}\right]^{0.5}\left[75+ \frac{1}{x^2}\right]^{0.5}= \left[85+ \frac{1}{x^2}\right]^{0.5}+ \left[75+ \frac{1}{x^2}\right]^{0.5})
.
Now, if you square both sides you will get rid of the square roots on the left but, because the square of a sum involves a "cross term", you will still have a square root on the right:
![\frac{x^2}{4}\left[85+ \frac{1}{x^2}\right]\left[75+ \frac{1}{x^2}\right]= \left[85+ \frac{1}{x^2}\right]+ 2\left[85+ \frac{1}{x^2}\right]^{0.5}\left[75+ \frac{1}{x^2}\right]^{0.5}+ \left[75+ \frac{1}{x^2}\right]](http://latex.codecogs.com/png.latex?\frac{x^2}{4}\left[85+ \frac{1}{x^2}\right]\left[75+ \frac{1}{x^2}\right]= \left[85+ \frac{1}{x^2}\right]+ 2\left[85+ \frac{1}{x^2}\right]^{0.5}\left[75+ \frac{1}{x^2}\right]^{0.5}+ \left[75+ \frac{1}{x^2}\right])
What you can do is move all except that "middle term", still involving square roots, to the left side of the equation and then square again. The good news is that you have now removed all square roots and have a polynomial equation. The bad news is that that polynomial equation will have degree 12!
Re: Can this equation be solved for F?
Wow, thanks. I had a feeling if it could be solved it would be something ridiculous like that!
