Solving Quadratic Equations with y=A/[B+(x-C)^2]

In summary: It's a law of physics that states the electric force between two charges is inversely proportional to the square of the distance between them.
  • #1
asca
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New poster has been reminded to show the Relevant Equations and their work on the solution
Homework Statement
Folks, I'm sure it is immediate to many of you, but the solution is not crossing my mind
Relevant Equations
Which phenomenon could be described by equation below?
y=A/[B+(x-C)^2]
 
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  • #2
Is there any reason it could not be an equation for resonance of some kind. Could also be a Cauchy density or something similar, although it would have to be normalized to be a density. In what context did it appear. This might help tell you something
 
  • #3
asca said:
Homework Statement:: Folks, I'm sure it is immediate to many of you, but the solution is not crossing my mind
Relevant Equations:: Which phenomenon could be described by equation below?

y=A/[B+(x-C)^2]
As a physical phenomenon, superficially it is dimensionally inconsistent. I.e. you have B added to the square of something. So it should help to think of B as also being the square of something.

(In future, please use thread titles that give some clue as to the subject matter.)
 
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  • #4
The only thing i could think of is movement of a particle in the curve that is described by $$y=\frac{A}{B+(x-C)^2}$$

This curve looks like some sort of parabola with peak at x=C , and with inverted edges, but still the equation suffers that is dimensionally inconsistent (the units on the left and right side of the equation don't match since x and y should me in the same units of length e.g meters).

To fix the units issue we should multiply the term ##(x-C)^2## by a constant ##\alpha## that will be equal to 1 but will have the proper units.
 
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  • #5
Thank you all for your attempts, and please forgive me if I posted the question in an inapproriate way. Unfortunately the question is the exact question the exam poses. They ask you to deduct which physical situation could be described by such kind of equation. And once you get to the answer, they ask you to describe the dimensions of A B and C and to talk about the physiscs involved.
I was thinking about something that has to do with the impedence of an LRC circuit, which is very similar to the denominator, or else some friction dumping in and harminc oscillator, but still I cannot imagine which real problem would imply that equation. Thank you all.
 
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  • #6
asca said:
Thank you all for your attempts, and please forgive me if I posted the question in an inapproriate way. Unfortunately the question is the exact question the exam poses. They ask you to deduct which physical situation could be described by such kind of equation. And once you get to the answer, they ask you to describe the dimensions of A B and C and to talk about the physiscs involved.
I was thinking about something that has to do with the impedence of an LRC circuit, which is very similar to the denominator, or else some friction dumping in and harminc oscillator, but still I cannot imagine which real problem would imply that equation. Thank you all.
Ok, but as I wrote it should help to express B as the square of something:
##\frac A{b^2+(x-C)^2}##
What does the sum of two squares make you think of?
 
  • #7
Gee I'm not getting it. Now I'm too focussed towards the RLC circuit I do not succeed in thinking about anything else...
 
  • #8
asca said:
Gee I'm not getting it. Now I'm too focussed towards the RLC circuit I do not succeed in thinking about anything else...
Think geometry. But it could also arise in circuits.
 
  • #9
You see, I keep seeing bsquared as R squared in the impedence formula. Geometry I obviously think about Pythagorean theorem. But I do not manage to imagine what are the sides of this triangle !
 
  • #10
maybe something that has to deal with inclined plane and a spring
 
  • #11
asca said:
maybe something that has to deal with inclined plane and a spring
Inverse square law?
But if you can make a case for RLC that's fine too. I'm sure there are several solutions to the question.
 
  • #12
No I can't make a case even for RLC. I do not see how I can get rid of the square root, unless I arbitarily square both sided of any eqautrion that comes to my mind...
 
  • #13
In a nutshell the problem seems to be: what is the ratio of something divided by the square of two other things that are 90 degrees apart form each other (in some space) e where one of those two has a symmetry point in C...
 
  • #14
Maybe I got it. What is the electric force bewteen a charge a q1 located in (C,0) and charge q2 located in (x,b)?
F= Kq1q2/[b^2+(x-C)^2] so y=F, A=kq1q2, B= b^2 and C is C. Bah ! What do you think?
 
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  • #15
asca said:
Maybe I got it. What is the electric force bewteen a charge a q1 located in (C,0) and charge q2 located in (x,b)?
F= Kq1q2/[b^2+(x-C)^2] so y=F, A=kq1q2, B= b^2 and C is C. Bah ! What do you think?
That'd do it.
 

FAQ: Solving Quadratic Equations with y=A/[B+(x-C)^2]

What is a quadratic equation?

A quadratic equation is a polynomial equation of the second degree, meaning it contains a variable raised to the power of two. It can be written in the form y = ax^2 + bx + c, where a, b, and c are constants and x is the variable.

How do you solve a quadratic equation with y=A/[B+(x-C)^2]?

To solve this type of quadratic equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Plug in the values for a, b, and c from the equation y = A/[B+(x-C)^2] and solve for x.

Can you solve a quadratic equation with complex solutions using this method?

Yes, the quadratic formula can be used to solve quadratic equations with complex solutions. The √ symbol in the formula represents the square root, so if the value inside the square root is negative, the solutions will be complex numbers.

Are there any other methods for solving quadratic equations?

Yes, there are other methods for solving quadratic equations such as factoring, completing the square, and graphing. However, the quadratic formula is a reliable method that can be used for any quadratic equation.

Can you use this method to solve quadratic equations with multiple variables?

Yes, the quadratic formula can be used to solve quadratic equations with multiple variables. Simply substitute the variables for the corresponding values and solve for the variable you are interested in finding.

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