Solving 0 in Physics E&M Equation

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The discussion revolves around solving the equation E = 0 = 10 / (1 - x)² + 5 / x² - 10 / (1 + x)² in a calculus-based E&M course. The original poster is uncertain about how to find the roots and notes that the solutions manual provides a numerical answer without clear steps. Suggestions include putting the fractions on a common denominator and using Newton's Method for numerical solutions. The poster expresses skepticism about the necessity of Newton's Method for a general physics course and plans to consult their professor for simpler alternatives. The conversation highlights the challenges of solving complex equations in introductory physics.
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I am working on a physics problem from a calc-based E&M (intro) course, and I came up with the following equation:

E = 0 = 10 / (1 - x)2 + 5 / x2 - 10 / (1 + x)2

I got here and worried it was a dead end, so I checked the solutions manual, and they have the exact same thing, and then say " therefore, the root is bla. So, ~equation~ = 0 => x = bla"

How did they solve this? =|

edit: The only attempt I would know how to make is solve one of the terms individually, and then solve the other two as a system, and hope that one of the two solutions coincide. However, this seems stupid because that basically leaves it all to chance.
 
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Hi anonymity! :smile:

Put the fractions on the same denominator and add them up. That's always a good first step!
 
Not a major point, but your title is screwy. You never "solve for zero." We know what zero is. You're solving for x, which is not known.
 
Yeah I know I read that as I was closing out the window to head to class..didn't have time or motivation enough to edit it, as it is irrelevant and clarified in my post.

Micro: thanks ill try that, seems like it could get very messy very quick though : I
 
anonymity said:
I am working on a physics problem from a calc-based E&M (intro) course, and I came up with the following equation:

E = 0 = 10 / (1 - x)2 + 5 / x2 - 10 / (1 + x)2

I got here and worried it was a dead end, so I checked the solutions manual, and they have the exact same thing, and then say " therefore, the root is bla. So, ~equation~ = 0 => x = bla"

How did they solve this? =|

edit: The only attempt I would know how to make is solve one of the terms individually, and then solve the other two as a system, and hope that one of the two solutions coincide. However, this seems stupid because that basically leaves it all to chance.
Did they give a formula, or a numerical answer? If they gave a formula, what is it?

If they give a formula they might have used the general solution of a quartic equation as given in http://en.wikipedia.org/wiki/Quartic_function , for example. If they gave a numerical answer they might have used Newton's Method.

RGV
 
It was a numerical value. I can't believe I didnt think of using Newton's method (i can't believe, either, that this actually showed up in a class other than basic single variable calculus...).

As far as if this is the way the book did it, I am not so sure. It seems unlikely that they would REQUIRE you to use something as obscure as Newton's method for a general physics course.

Regardless, assuming Newton's method doesn't fail and bounce all around the number plane, it will work. I will ask my professor tomorrow if there's a less tedious/more basic way to solve it.

Thanks for your help Ray.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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