Does anyone wish to solve the following quadradic equation for x₂?

x₂²(1-(v/c)²) - x₂(x₀+(v/c)y) + x₀²-2x₀(v/c)y = 0

What a mess. I have it in the standard form [-b +/- sqrt(b^2 -4AC)]/2A but all the squaring and multiplying and square-rooting... ugh.

Just use the quadratic formula.

I wanted to use a symbolic solver but can't locate my calculator :/

The quadradic forumla is a bigger mess, try it.

I did. A mess.

Any online symbolic calculators?

x₀ + sqrt[(x₀+(v/c)y)² - 4(1-(v/c)²)(x₀²-2x₀(v/c)y)]
x₂ = ---------------------------------------------------------------------------
(2-2(v/c)²)

Superluminal's solution is fine. You can use that.

I already had that, I don't like that :( You're mean James R.

What's the relevance of the quadratic equation, anyway?

Yeah, anyway?

This is a math forum is it not? I posted a math question did I not? Does there have to be any meaning? ;)

No, that's fine Aer. Whatever.

I see your little winky face. What devilishness are you plotting?

x₂ should be the apparent distance between two objects taking into account special relativity and photon travel delay.

What are x₀, y, c and v?

x₀ is not the proper length. (you probably won't like the notation).

Anyway, x₀ is the distance between 2 objects moving at the same speed v relative to the observer as special relativity would give. ( i.e. x₀=l=l₀/γ ) This can be measured directly if the observer is directly in the middle of the two observers such that photons take the same amount of time to travel to the observer. However, for any other case, there is going to be a photon travel time delay which can be calculated if y is the distance between the objects and observer (perpendicular distance to travel of the objects that is). c is the speed of light.

Furthermore, I should stipulate, that x₂ was a special case when object A is directly perpendicular to the observer because that is the case I am interested in.

Like this:

A<-----x₂----->B -->v



O

If you don't think I approached it right, here were all my steps:

t₀ is the point in time when O would be directly between A and B.

t₂ is when object A is directly perpendicular to O.

t₁ is when object B gives off a photon that O would measure at the same time the photon from A as given off at time t₂ would reach O.

t₃ is when the photons from A & B, both reach O.

t₂ = x₀/(2V)

t₃ = t₂ + y/c

t₃ = t₁ + sqrt(y²+x₂²)/c

x₂ = x₀/2 + vt₁

which lead to the equation in the OP.

I am almost certain that there was an error in going from the 4 equations above to the equation in the OP but I am too tired to find out where I screwed up.

I wish sciforums supported LaTeX.

OK - the equation in the OP is wrong.

Anyway, the solution is:

x₂ = γ²(x₀+(v/c)*(y-sqrt(y²+x₀²+2x₀y(v/c))))


I doubt it simplifies much more.

And I believe it is correct because x₂ should approach x₀ as v goes to 0 which clearly is the case in the equation above.