# Solving $4t+{2\over t-3}+{2\over3}=0$ [duplicate]

Possible Duplicate:
How would I rearrange this equation to make the subject $t$?

The scenario is as follows:

A sub-atomic particle is travelling in a straight line through a tubular cloud-chamber that is 28 cm long. The particle is subjected to an electromagnetic field that reverses the direction of the particle so that it disappears from the cloud-chamber. Almost instantaneously with the disappearance of the first particle, a second particle enters the chamber from the other direction. This particle is subjected to the same electromagnetic field so that its direction is also reversed and it disappears from the cloud-chamber after a period of time along the same trajectory as the first particle would have done if the electromagnetic field had not been applied. It is possible to model approximately these events with the function

$$s(t) = 4t + \frac2{t-3} + \frac23$$

where s represents the position of either particle in the tubular cloud-chamber measured in centimetres, while t represents the time in nano-seconds.

The question is:

When does the first particle actually leave the cloud-chamber according to the model?

[Hint: consider s(t) = 0]

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## marked as duplicate by Gerry Myerson, Kannappan Sampath, Henning Makholm, Asaf Karagila, Zev ChonolesMar 4 '12 at 9:16

In facts, it looks like at $t=2.8$ the first particle is still inside the chamber, but moving significantly faster than the speed of light. – Henning Makholm Mar 3 '12 at 3:12
@Christy McGrory: You are welcome. You had carried out the multiplications successfully, and had more or less arrived at $12t^2=34t=0$. It would have been overkill (but possible) to use the Quadratic Formula on that, since $12t^2-34t$ factors so simply. – André Nicolas Mar 3 '12 at 5:05
You were given the hint (or command?) to find the time $t$ when $s(t)=0$. We will do that, although the relevance to the problem as stated is not entirely clear. So we want to solve the equation $$4t + \frac{2}{t-3} + \frac{2}{3}=0.$$ Multiply through by $3(t-3)$. This is legitimate, since we cannot have $t=3$. We arrive at the equivalent equation $$(3)(t-3)(4t)+(3)(2)+(t-3)(2)=0.$$ This simplifies to $$12t^2-34t=0, \quad\text{and then to}\quad t(12t-34)=0.$$ Since the solution $t=0$ is irrelevant. we arrive at the equation $12t-34=0$, whose solution is $t=\dfrac{34}{12}$.